If one of the 1008 subjects is randomly selected, find the probability that the person chosen is a woman given that the person is a light smoker. Round to the nearest thousandth.
Answers
Answer:
The probability is 0.4841.
Step-by-step explanation:
The provided table is:
From above table, it is known that
Number of subjects are 1008.
The probability that the person chosen is a woman given that the person is a light smoker can be calculated as:
[tex]P(Woman| Light smoker)=\frac{P(Woman and light smoker)}{P(Light smoker)} \\P(Woman| Light smoker)= \frac{\frac{76}{1008} }{\frac{157}{1008} } = 0.4841[/tex]
Thus, required probability is 0.4841.
The probability that a randomly selected person is a woman given that the person is a light smoker is [tex]\[ 0.400} \][/tex]
To find the probability that a randomly selected person is a woman given that the person is a light smoker, we need to use conditional probability. The formula for conditional probability [tex]\( P(A|B) \)[/tex] is:
[tex]\[P(A|B) = \frac{P(A \cap B)}{P(B)}\][/tex]
Where:
[tex]\( P(A|B) \)[/tex] is the probability of event [tex]\( A \)[/tex] occurring given that [tex]\( B \)[/tex] has occurred.
[tex]\( P(A \cap B) \)[/tex] is the probability of both events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] occurring.
[tex]\( P(B) \)[/tex] is the probability of an event [tex]\( B \)[/tex] occurring.
Let's define the events:
[tex]\( A \)[/tex] : The person chosen is a woman.
[tex]\( B \)[/tex] : The person chosen is a light smoker.
We need the number of light smokers and the number of women who are light smokers. Suppose we have the following data:
Total number of subjects: 1008
Number of light smokers: [tex]\( n_{\text{light smokers}} \)[/tex]
A number of women who are light smokers: [tex]\( n_{\text{women and light smokers}} \)[/tex]
Given that the number of women who are light smokers is [tex]\( n_{\text{women and light smokers}} \)[/tex] and the total number of light smokers is [tex]\( n_{\text{light smokers}} \)[/tex], the probability can be calculated as follows:
[tex]\[P(\text{woman | light smoker}) = \frac{n_{\text{women and light smokers}}}{n_{\text{light smokers}}}\][/tex]
If we don't have the exact numbers, we'll need those to calculate the probability. However, let's assume the following values (hypothetically for the purpose of illustration):
Total number of light smokers: 150
Number of women who are light smokers: 60
The probability that a randomly selected person is a woman given that the person is a light smoker is:
[tex]\[P(\text{woman | light smoker}) = \frac{60}{150} = 0.4\][/tex]
Rounding to the nearest thousandth:
[tex]\[0.4 = 0.400\][/tex]
Related Questions
According to a July 31, 2013, posting on cnn subsequent to the death of a child who bit into a peanut, a 2010 study in the journal Pediatrics found that 8% of children younger than 18 in the United States have at least one food allergy. Among those with food allergies, about 39% had a history of severe reaction. a. If a child younger than 18 is randomly selected, what is the probability that he or she has at least one food allergy and a history of severe reaction? b. It was also reported that 30% of those with an allergy in fact are allergic to multiple foods. If a child younger than 18 is randomly selected, what is the probability that he or she is allergic to multiple foods?
Answers
Answer:
a. 0.0312 or 3.12%
b. 0.024 or 2.40%
Step-by-step explanation:
a. The probability that randomly selected child has at least one food allergy and a history of severe reaction is determined by the probability of having a food allergy (8%) multiplied by the probability of having a history of severe reaction (39%):
[tex]P = 0.08*0.39\\P=0.0312 = 3.12\%[/tex]
b. The probability that randomly selected child is allergic to multiple foods is determined by the probability of having a food allergy (8%) multiplied by the probability of being allergic to multiple foods (30%):
[tex]P = 0.08*0.30\\P=0.024 = 2.40\%[/tex]
According to the Bureau of Labor Statistics it takes an average of 16 weeks for young workers to find a new job. Assume that the probability distribution is normal and that the standard deviation is two weeks. What is the probability that 20 young workers average less than 15 weeks to find a job?
Answers
Answer:
1.25% probability that 20 young workers average less than 15 weeks to find a job
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 16, \sigma = 2, n = 20, s = \frac{2}{\sqrt{20}} = 0.4472[/tex]
What is the probability that 20 young workers average less than 15 weeks to find a job?
This is the pvalue of Z when X = 15. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{15 - 16}{0.4472}[/tex]
[tex]Z = -2.24[/tex]
[tex]Z = -2.24[/tex] has a pvalue of 0.0125.
1.25% probability that 20 young workers average less than 15 weeks to find a job
Lance bought n notebooks that cost $0.75 each and p pens that cost $0.55 each. A 6.25% sales tax will be applied to the total cost. Which expression represents the total amount Lance paid, including tax?
Answers
Answer: 0.796875n + 0.584375p
Step-by-step explanation:
Lance bought n notebooks that cost $0.75 each. This means that the total cost of n notebooks would be $0.75n
Lance also bought p pens that cost $0.55 each. This means that the total cost of p pens would be $0.55p
The total cos of n notebooks and p pens is
0.75n + 0.55p
A 6.25% sales tax will be applied to the total cost. This means the amount of tax paid would be
0.0625(0.75n + 0.55p)
= 0.046875n + 0.034375p
Therefore, the expression that represents the total amount Lance paid, including tax is
0.75n + 0.55p + 0.046875n + 0.034375p
= 0.796875n + 0.584375p
Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study27 has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is 0.183. The standard error for this estimate is 0.016.
a.Give notation for the quantity being estimated, and define any parameters used.
b.Give notation for the quantity that gives the best estimate, and give its value.
c.Give a 95% confidence interval for the quantity being estimated. Interpret the interval in context.
Answers
Answer:
a) [tex]\mu[/tex]
b) [tex]\bar{x}[/tex]
c) (0.152, 0.214)
Step-by-step explanation:
We are given the following in the question:
Sample mean = 0.183 ppm
Standard error = 0.016
a) quantity being estimated
We have to estimate the population mean.
Notation for population mean:
[tex]\mu[/tex]
b) Best estimate for population mean is the sample mean
Notation for sample mean:
[tex]\bar{x}[/tex]
The point estimate for population mean is
[tex]\mu = \bar{x} = 0.183[/tex]
c) 95% confidence interval
[tex]\bar{x}\pm z_{critical}(\text{Standard Error})[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
Putting values, we get,
[tex]0.183 \pm 1.96(0.016)\\=0.183\pm 0.03136\\=(0.15164, 0.21436)\\\approx (0.152, 0.214)[/tex]
Thus, the 95% confidence interval is:
(0.152, 0.214)
Interpretation:
After treatment with moose drool, we are 95% certain or 95% confident that the interval (0.152, 0.214) contains the true mean of the population that is the mean level of the toxin ergovaline on the grass.
The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The best estimate for this quantity is 0.183 ppm. The 95% confidence interval for this estimate is (0.15164, 0.21436) ppm.
Explanation:a. The quantity being estimated is the mean level of the toxin ergovaline on the grass after treatment with moose drool. The parameter used is the standard error, which measures the variability of the estimate.
b. The notation for the quantity that gives the best estimate is the mean level of the toxin ergovaline after treatment with moose drool, denoted as μ.
c. To calculate a 95% confidence interval, we need to determine the margin of error. Since the standard error is given as 0.016, the margin of error is 1.96 times the standard error, which is approximately 0.03136. The 95% confidence interval is then calculated by subtracting and adding the margin of error from the mean level of the toxin ergovaline after treatment with moose drool, resulting in an interval of (0.15164, 0.21436). This means that we are 95% confident that the true mean level of the toxin ergovaline after treatment with moose drool is between 0.15164 and 0.21436 ppm.
Learn more about Estimating mean level of toxin ergovaline here:https://brainly.com/question/32478139
#SPJ3
The random variable x has a normal distribution with standard deviation 21. It is known that the probability that x exceeds 160 is .90. Find the mean mu of the probability distribution.
Answers
Answer:
Mean, [tex]\mu[/tex] = 133.09
Step-by-step explanation:
We are given that the random variable x has a normal distribution with standard deviation 21,i.e;
X ~ N([tex]\mu,\sigma = 21[/tex])
The Z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ (0,1)
Also, it is known that the probability that x exceeds 160 is 0.90 ,i.e;
P(X > 160) = 0.90
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{160-\mu}{21}[/tex] ) = 0.90
From the z% table we find that at value of x = -1.2816, the value of
P(X > 160) is 90%
which means; [tex]\frac{160-\mu}{21}[/tex] = -1.2816
160 - [tex]\mu[/tex] = 21*(-1.2816)
[tex]\mu[/tex] = 160 - 26.914 = 133.09
Therefore, mean [tex]\mu[/tex] of the probability distribution is 133.09.
I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and provide statistical explanations. Please help me and answer these before 9: 15 in the morning EST on March 3rd, 2020.
Answers
Answer:
question 1[tex]n^{2} - 20n -96 = 0[/tex]
use product and sum method
product = -96
sum = -20
numbers needed = ( -24 , 4)
n - 24 = 0
n + 4 = 0
hence n = 24 and n = -4
Question 2
[tex]x^{2} + 12 x = 48[/tex]
in the form [tex]ax^{2} +bx +c = 0[/tex]
= [tex]x^{2} +12x - 48[/tex]
make use of the formula :
[tex]\frac{-b+-\sqrt{b^{2} -4ac} }{2a}[/tex]
replace values to make 2 equations :
1.[tex]\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = 3.17
2.[tex]\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}[/tex] = -15.2
hence x = 3.17 and x = -15.2
Question 3
[tex]x^{2} -14x+40=0[/tex]
use product and sum method
product = 40
sum = -14
numbers needed = (-10 , -4)
x - 10 = 0
x - 4 = 0
hence x = 10 and x = 4
Question 4
[tex]5b^{2} -20b-18 = 7[/tex]
in the form [tex]ax^{2} +bx +c = 0[/tex]
this becomes [tex]5b^{2} -20b-18-7[/tex]
= [tex]5b^{2} -20b-25[/tex]
can simplify by 5
= [tex]b^{2} -4b-5 =0\\[/tex]
use product and sum method
product = -5
sum = -4
numbers needed (-5 , 1)
b-5 = 0
b + 1 = 0
hence b = 5 and b = -1
Answer:
Step-by-step explanation:
1) n² - 20n - 96 = 0
n² - 20n + (- 20/2)² = 96 + (- 20/2)²
(n - 10)² = 96 + 100
(n - 10)² = 196
Taking square root of both sides
n - 10 = √196 = 14
n = 14 + 10
n = 24
2) x² + 12x = 48
x² + 12x + (12/2)² = 48 + (12/2)²
(x + 6)² = 48 + 36 = 84
Taking square root of both sides,
x + 6 = 9.2
x = 9.2 - 6
x = 3.2
3) x² - 14x + 40 = 0
x² - 14x = - 40
x² - 14x + (- 14/2)² = - 40 + (- 14/2)²
(x - 7)² = - 40 + 49 = 9
Taking square root of both sides,
x - 7 = 3
x = 3 + 7
x = 10
4) 5b² - 20b - 18 = 7
5b² - 20b = 7 + 18
5b² - 20b = 25
Dividing both sides by 5, it becomes
b² - 4b = 5
b² - 4b + (-4/2)² = 5 + (-4/2)²
(b - 2)² = 5 + 4 = 9
Taking square root of both sides
b - 2 = 3
b = 3 + 2
b = 5
Popcorn kernels pop independently (i.e. unimolecularly). For one brand at constant temperature, 7 kernels pop in 10 seconds when 180 kernels are present. After 75 kernels have popped, how many kernels will pop in 10 seconds? (Your answer may include fractions of a kernel).
Answers
Answer:
In 10 seconds, after 75 kernels have popped, they should pop 161/36 kernels.
Step-by-step explanation:
You can solve this problem with a rule of three, if you remove 75 kernels, you have 115 left, and in 10 seconds you may expect a proportional amount of popcorn kernels dropping.
If for 180 kernels 7 drop, then for 115 kernels the amount that drops is
7/180 * 115 = 161/36 popcorn kernels.
5)
Solve the equation for x, in the simplest form.
2/3x = 2/5
A) x = 2/5
B) x = 3/5
C) x = 1/2
D) x = 5/6
Answers
Answer:
B x = 3/5
Step-by-step explanation:
2/3 x = 2/5
Multiply each side by 15 to get rid of the fractions
15(2/3 x )= 2/5*15
10x = 6
Divide each side by 10
10x/10 =6/10
x = 6/10
We can simplify this fraction. Divide each side by 2
x = 3/5
Consider a value to be significantly low if its z score less than or equal to minus 2 or consider a value to be significantly high if its z score is greater than or equal to 2. A test is used to assess readiness for college. In a recent year, the mean test score was 21.6 and the standard deviation was 5.4. Identify the test scores that are significantly low or significantly high. What test scores are significantly low?
Answers
Answer:
The scores that are less than or equal to 10.8 are considered significantly low.
The scores that are greater than or equal to 32.4 are considered significantly high.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 21.6
Standard Deviation, σ = 5.4
We are given that the distribution of score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Significantly low score:
[tex]z \leq -2\\z = \displaystyle\frac{x-21.6}{5.4} \leq -2\\\\\displaystyle\frac{x-21.6}{5.4} \leq -2\\\\x\leq -2(5.4) + 21.6\\\Rightarrow x \leq 10.8[/tex]
Thus, scores that are less than or equal to 10.8 are considered significantly low.
Significantly high score:
[tex]z \geq 2\\z = \displaystyle\frac{x-21.6}{5.4} \geq 2\\\\\displaystyle\frac{x-21.6}{5.4} \geq 2\\\\x\geq 2(5.4) + 21.6\\\Rightarrow x \geq 32.4[/tex]
Thus, scores that are greater than or equal to 32.4 are considered significantly high.
A 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall. What is the 5-day growth factor for the height of the sunflower
Answers
Answer:
The growth factor is approximately 0.11 or 11%.
Step-by-step explanation:
We have been given that a 8-inch tall sunflower is planted in a garden and the height of the sunflower increases exponentially. 5 days after being planted the sunflower is 13.4805 inches tall.
We will use exponential growth formula to solve our given problem.
[tex]y=a\cdot (1+r)^x[/tex], where,
y = Final value,
a = Initial value,
r = Growth rate in decimal form,
x = Time.
Upon substituting initial value [tex]a=8[/tex], [tex]x=5[/tex] and [tex]y=13.4805[/tex], we will get:
[tex]13.4805=8\cdot(1+r)^5[/tex]
[tex]8\cdot(1+r)^5=13.4805[/tex]
[tex]\frac{8\cdot(1+r)^5}{8}=\frac{13.4805}{8}[/tex]
[tex](1+r)^5=\frac{13.4805}{8}[/tex]
Now, we will take 5th root of both sides of equation as:
[tex]\sqrt[5]{(1+r)^5} =\sqrt[5]{\frac{13.4805}{8}}[/tex]
[tex]1+r =\sqrt[5]{1.6850625}[/tex]
[tex]1+r =1.1100005724234515[/tex]
[tex]1-1+r =1.1100005724234515-1[/tex]
[tex]r=0.1100005724234515[/tex]
[tex]r\approx 0.11[/tex]
Therefore, the growth factor is approximately 0.11 or 11%.
The 5-day growth factor of the sunflower, which grows exponentially, is calculated by dividing the final height (13.4805 inches) by the initial height (8 inches) giving a growth factor of approximately 1.6850625.
Explanation:The 5-day growth factor of the sunflower that increases exponentially can be calculated by dividing the final height by the initial height. The formula is:
5-day growth factor = final height / initial height
To calculate the 5-day growth factor, we substitute the heights into the formula. We have:
5-day growth factor = 13.4805 inches (final height) / 8 inches (initial height)
So, 5-day growth factor = 1.6850625. This means that the height of the sunflower is multiplied by approximately 1.685 each day for 5 days, which accounts for the exponential growth.
Learn more about growth factor here:https://brainly.com/question/32122796
#SPJ3
A factorial experiment was designed to test for any significant differences in the time needed to perform English to foreign language translations with two computerized language translators. Because the type of language translated was also considered a significant factor, translations were made with both systems for three different languages: Spanish, French, and German. Use the following data for translation time in hours.
Language
Spanish French German
System 1 9 12 13
13 16 17
System 2 9 15 19
13 17 25
Test for any significant differences due to language translator system (Factor A), type of language (Factor B), and interaction. Use = .05.
Answers
Answer:
See the attached pictures for answer.
Step-by-step explanation:
See the attached pictures for explanation.
Answer:
Step-by-step explanation:
We have a factorial experiment with Factors A and B, levels a=2, b=3, and n=2 observations.
It's necessary to apply a Two-Factor with replication ANOVA by using Excel or calculating by hand.
1. Sum of Squares could be calculated with the following formulas:
[tex]SS_{T}=\sum\limits^a_{i=1}\sum\limits^b_{j=1}\sum\limits^n_{k=1} {(y_{ijk}-y_{...} ^{2} )}\\SS_{A}=bn\sum\limits^a_{i=1} {(y_{i..}^{2}) -Ny_{...} ^{2} }\\SS_{B}=an\sum\limits^b_{j=1} {(y_{.j.}^{2}) -Ny_{...} ^{2} }\\SS_{AB}=n\sum\limits^a_{i=1}\sum\limits^b_{j=1} {(y_{ij.}^{2}) -Ny_{...} ^{2} -SC_{A} -SC_{B}-SC_{AB}}[/tex]
2. Calculate Degrees of Freedom
Factor A: [tex]a-1[/tex]
Factor B: [tex]b-1[/tex]
Interaction: [tex](a-1)(b-1)[/tex]
Within: [tex]ab(n-1)[/tex]
Total: [tex]abn-1[/tex]
3. Mean Square:
Factor A: [tex]MS_{A}=\frac{SS_{A}}{a-1}[/tex]
Factor B: [tex]MS_{B}=\frac{SS_{B}}{b-1}[/tex]
Interaction: [tex]MS_{AB}=\frac{SS_{AB}}{(a-1)(b-1)}[/tex]
Within:[tex]MS_{E}=\frac{SS_{E}}{ab(n-1)}[/tex]
4. Estimate F and P-value
[tex]F_{A}=\frac{SS_{A} }{SS_{E}}[/tex]
[tex]F_{B}=\frac{SS_{B} }{SS_{E}}[/tex]
[tex]F_{AB}=\frac{SS_{AB} }{SS_{E}}[/tex]
Open the attachment to see the results.
5. Analysis of P-values
Factor A: P-value (0.1280 )>(0.05)
Factor B: P-value (0.0315 )<(0.05)
Interaction: P-value (0.2963 )<(0.05)
P-value of Factor B is less than 0.05, then type of language is insignifficant.
P-value of Factor A and interaction are greater than 0.05, then the translator system and the interaction of both factors are signifficant.
There are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class project. How divisions are possible?
Answers
Answer:
27720
Step-by-step explanation:
Given that there are 12 students in a graduate class. The students are to be divided into three groups of 3, 4, and 5 members for a class project.
From 12 students 3 students for group I can be selected in 12C3 ways.
Now from remaining 9, 4 students can be selected for II group in 9C4 ways
The remaining 5 have to be placed in III group.
Hence possible divisions for grouping the 12 students in the class into three groups
= 12C3 *9C4
= [tex]=220*126=27720[/tex]
Final answer:
There are 27,720 possible divisions of the 12 students into three groups of 3, 4, and 5 members.
Explanation:
To determine the number of possible divisions, we need to find the number of ways to choose 3 students from 12 for the first group, then 4 students from the remaining 9 for the second group, and finally 5 students from the remaining 5 for the third group.
Using the combination formula, the number of ways can be calculated as:
Number of ways = C(12, 3) * C(9, 4) * C(5, 5)
C(12, 3) = 12! / (3! * (12-3)!) = 220
C(9, 4) = 9! / (4! * (9-4)!) = 126
C(5, 5) = 5! / (5! * (5-5)!) = 1
Therefore, the number of possible divisions is 220 * 126 * 1 = 27,720.
Consider an experiment with sample space S 5 50, 1, 2, 3, 4, 5, 6, 7, 8, 96 and the events A 5 {0, 2, 4, 6, 8} B 5 {1, 3, 5, 7, 9} C 5 {0, 1, 2, 3, 4} D 5 {5, 6, 7, 8, 9} Find the outcomes
Answers
Answer with Step-by-step explanation:
S={0,1,2,3,4,5,6,7,8,9}
A={0,2,4,6,8}
B={1,3,5,7,9}
C={0,1,2,3,4}
D={5,6,7,8,9}
a.A'=S-A
A'={0,1,2,3,4,5,6,7,8,9}-{0,2,4,6,8}
A'={1,3,5,7,9}
b.C'=S-C
C'={0,1,2,3,4,5,6,7,8,9}-{0,1,2,3,4}
C'={5,6,7,8,9}
c.D'=S-D
D'={0,1,2,3,4,5,6,7,8,9}-{5,6,7,8,9}
D'={0,1,2,3,4}
d.[tex]A\cup B=[/tex]{0,2,4,6,8}[tex]\cup[/tex]{1,3,5,7,9}
[tex]A\cup B=[/tex]{0,1,2,3,4,5,6,7,8,9}=S
e.[tex]A\cup C[/tex]={0,2,4,6,8}[tex]\cup[/tex]{0,1,2,3,4}
[tex]A\cup C[/tex]={0,1,2,3,4,6,8}
f.[tex]A\cup D[/tex]={0,2,4,6,8}[tex]\cup[/tex]{5,6,7,8,9}
[tex]A\cup D[/tex]={0,2,4,5,6,7,8,9}
The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a 90% confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a 90% confidence interval for the proportion of college seniors who carry a credit card balance from month to month.c. Explain why the two 90% confidence intervals from Parts (a) and (b) are not the same width.
Answers
Answer:
Step-by-step explanation:
Hello!
There are two variables of interest:
X₁: number of college freshmen that carry a credit card balance.
n₁= 1000
p'₁= 0.37
X₂: number of college seniors that carry a credit card balance.
n₂= 1000
p'₂= 0.48
a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.
The formula for the interval is:
p'₁±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.648[/tex]
0.37±1.648*[tex]\sqrt{\frac{0.37*0.63}{1000} }[/tex]
0.37±1.648*0.015
[0.35;0.39]
With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.
b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648
The formula for this interval is
p'₂±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }[/tex]
0.48±1.648* [tex]\sqrt{\frac{0.48*0.52}{1000} }[/tex]
0.48±1.648*0.016
[0.45;0.51]
With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.
c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.
Freshmen: [tex]\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015[/tex]
Seniors: [tex]\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016[/tex]
The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.
The confidence interval will be widest when [tex]\( \hat{p} = 0.5 \)[/tex] and narrowest when [tex]\( \hat{p} \)[/tex] is close to 0 or 1. In this case, the proportion for seniors is closer to 0.5 than the proportion for freshmen, resulting in a slightly wider confidence interval for seniors.
To construct 90% confidence intervals for the proportions of college freshmen and seniors who carry a credit card balance from month to month, we will use the formula for a confidence interval for a proportion:
[tex]\[ \text{Confidence Interval} = \hat{p} \pm Z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \][/tex]
where:
- [tex]\( \hat{p} \)[/tex] is the sample proportion,
- [tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level (for 90% confidence, the Z-score is approximately 1.645),
- \( n \) is the sample size.
For part (a), we have a sample proportion [tex]\( \hat{p}_{\text{freshmen}} = 0.37 \) and a sample size \( n_{\text{freshmen}} = 1000 \)[/tex]. The 90% confidence interval for the proportion of college freshmen is calculated as follows:
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.37(1 - 0.37)}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.37 \times 0.63}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \sqrt{\frac{0.2331}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 1.645 \times 0.0153 \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = 0.37 \pm 0.0252 \][/tex]
[tex]\[ \text{CI}_{\text{freshmen}} = (0.3448, 0.3952) \][/tex]
For part (b), we have a sample proportion [tex]\( \hat{p}_{\text{seniors}} = 0.48 \) and a sample size \( n_{\text{seniors}} = 1000 \)[/tex]. The 90% confidence interval for the proportion of college seniors is calculated as follows:
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.48(1 - 0.48)}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.48 \times 0.52}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \sqrt{\frac{0.2496}{1000}} \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 1.645 \times 0.0158 \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = 0.48 \pm 0.0260 \][/tex]
[tex]\[ \text{CI}_{\text{seniors}} = (0.4540, 0.5060) \][/tex]
For part (c), the reason why the two 90% confidence intervals from Parts (a) and (b) are not the same width is due to the different sample proportions[tex]\( \hat{p} \).[/tex] The width of a confidence interval for a proportion depends on the standard deviation of the sampling distribution of the proportion, which is given by [tex]\( \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)[/tex]. Since the sample proportions for freshmen and seniors are different (0.37 for freshmen and 0.48 for seniors), the standard deviations will also be different, leading to different widths for the confidence intervals.
Suppose that a random sample of size 36 is to be selected from a population with mean 43 and standard deviation 6. What is the approximate probability that X will be more than 0.5 away from the population mean?
Answers
Answer:
61.70% approximate probability that X will be more than 0.5 away from the population mean
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 36, \sigma = 6, n = 36, s = \frac{6}{\sqrt{36}} = 1[/tex]
What is the approximate probability that X will be more than 0.5 away from the population mean?
This is the probability that X is lower than 36-0.5 = 35.5 or higher than 36 + 0.5 = 36.5.
Lower than 35.5
Pvalue of Z when X = 35.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{35.5 - 36}{1}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085.
30.85% probability that X is lower than 35.5.
Higher than 36.5
1 subtracted by the pvalue of Z when X = 36.5. SO
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{36.5 - 36}{1}[/tex]
[tex]Z = 0.5[/tex]
[tex]Z = 0.5[/tex] has a pvalue of 0.6915.
1 - 0.6915 = 0.3085
30.85% probability that X is higher than 36.5
Lower than 35.5 or higher than 36.5
2*30.85 = 61.70
61.70% approximate probability that X will be more than 0.5 away from the population mean
To determine the probability that a sample mean is more than 0.5 away from the population mean, we calculate the z-score for the sample mean being 0.5 above or below the population mean, look up the cumulative probability for this z-score, double it to account for both tails, and subtract from 1.
To find the probability that a sample mean is more than 0.5 away from the population mean, we can use the concept of a sampling distribution. The central limit theorem (CLT) tells us that for a sample of size 36 (which is sufficiently large), the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the original population distribution.
Given the population mean (μ) is 43 and the population standard deviation (σ) is 6, the standard error of the mean (SEM) can be calculated as σ divided by the square root of the sample size (n), which in this case is 6/√36 = 1. The z-score for a value A that is 0.5 away from the mean (either 43.5 or 42.5) can be calculated using the formula (A-μ)/SEM. Thus, the z-score is (42.5-43)/1 = -0.5 or (43.5-43)/1 = 0.5.
Using the standard normal distribution table or a calculator for the cumulative distribution function, we can find the probability of a z-score being less than -0.5 or greater than 0.5. Since the distribution is symmetric, we can simply look up the probability of z < -0.5, then double it (to account for both tails of the distribution) and subtract from 1 to get the probability that the sample mean is more than 0.5 away from 43.
This Excel file Undergrad Survey shows the data resulting from a survey of 50 undergraduate students at Clemson University. Majors of students in the survey are accounting (A), economics and finance (EF), management (M), marketing (MR), computer information systems (IS), other (O), and undecided (UN). "Number of affiliations" is the number of social networking sites at which the student is registered; "Spending" is the amount spent on textbooks for the current semester. The other variables are self-explanatory.
We will assume that this sample is a representative sample of all Clemson undergraduates. Use Excel or statcrunch to make a histogram of GPA to verify that the distribution of GPA can be approximated by the N(3.12, 0.4) normal model.
Question 1. The School of Business at Clemson has created a rigorous new International Business Studies major to better prepare their students for the global marketplace. A GPA of 3.69 or higher is required for a Clemson undergraduate to change his/her major to International Business Studies. What is the probability that a randomly selected Clemson undergraduate has a GPA of at least 3.69? (Use 4 decimal places in your answer).
Question 2. To attract high-quality current Clemson undergraduates into the new International Business Studies major, scholarships in International Business Studies will be offered to a Clemson undergraduate if his/her GPA is at or above the 95.54th percentile. What is the minimum GPA required to meet this criterion?
The GPA Values:
2.38
2.42
2.45
2.50
2.60
2.61
2.65
2.67
2.74
2.75
2.75
2.76
2.80
2.87
2.88
2.91
2.92
2.93
2.94
3.00
3.02
3.09
3.10
3.11
3.13
3.14
3.18
3.19
3.20
3.21
3.22
3.23
3.24
3.26
3.28
3.33
3.34
3.43
3.44
3.48
3.50
3.55
3.62
3.62
3.63
3.71
3.72
3.77
3.85
4.00
Answers
Answer:
Question 1:
Here total number of students GPA is given = 50
Number of people who have GPA more than 3.69 = 5
Therefore,
Pr(That a student has GPA more than 3.69) = 5/50 = 0.1
Here X ~ NORMAL (3.12, 0.4)
so Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4)
Z = (3.69 - 3.12)/ 0.4 = 1.425
Pr(X > 3.69) = Pr(X > 3.69 ; 3.12 ; 0.4) = 1 - Pr(Z < 1.425) = 1 - 0.9229 = 0.0771
Question 2
Here GPA would be above 95.54th percentile
so as per Z table relative to that percentile is = 1.70
so Z = (X - 3.12)/ 0.4 = 1.70
X = 3.12 + 0.4 * 1.70 = 3.80
so any person with GPA above or equal to 3.80 is eligible for that.
If Brazil has a total of 150 bird species and you happen to catch two species in an observational experiment using mist nets, what is the total number of possible combinations of species that you captured
Answers
Answer:
The total number of possible combinations of species that you captured is 11,175.
Step-by-step explanation:
The order that you captured the species is not important. For example, capturing a bird of specie A then a bird of specie B is the same outcomes as capturing a bird of specie B then a bird of specie A. So the combinations formula is used to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, we have that:
Combinations of 2 species from a set of 150. So
[tex]C_{150,2} = \frac{150!}{2!148!} = 11,175[/tex]
The total number of possible combinations of species that you captured is 11,175.
In a population of 200,000 people, 40,000 are infected with a virus. After a person becomes infected and then recovers, the person is immune (cannot become infected again). Of the people who are infected, 5% will die each year and the others will recover. Of the people who have never been infected, 35% will become infected each year. How many people will be infected in 4 years? (Round your answer to the nearest whole number.)
Answers
After 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.
In a population of 200,000 people, 40,000 are initially infected with a virus. Each year, 5% of the infected individuals die, and 25% of the uninfected population becomes infected. The population dynamics over the next 4 years can be analyzed step by step:
Year 0:
Total Population: 200,000
Infected: 40,000
Uninfected: 160,000
Year 1:
Infected: 40,000 + (25% of 160,000) = 40,000 + 40,000 = 80,000
Uninfected: 160,000 - 40,000 = 120,000
Deaths: 5% of 80,000 = 4,000
Recoveries: 95% of 80,000 = 76,000
Year 2:
Infected: (25% of 120,000) = 30,000
Uninfected: 120,000 - 30,000 = 90,000
Deaths: 5% of 30,000 = 1,500
Recoveries: 95% of 30,000 = 28,500
Year 3:
Infected: (25% of 90,000) = 22,500
Uninfected: 90,000 - 22,500 = 67,500
Deaths: 5% of 22,500 = 1,125
Recoveries: 95% of 22,500 = 21,375
Year 4:
Infected: (25% of 67,500) = 16,875
Uninfected: 67,500 - 16,875 = 50,625
Deaths: 5% of 16,875 = 844
Recoveries: 95% of 16,875 = 16,031
In summary, after 4 years, approximately 16,875 people will be infected due to the virus, considering deaths, recoveries, and new infections.
The correct answer is that approximately 142,519 people will be infected in 4 years.
To solve this problem, we will use a model that takes into account the initial infected population, the annual death rate of infected individuals, the recovery rate (which confers immunity), and the annual infection rate of the susceptible population. We will calculate the number of infected individuals at the end of each year and then sum them up to find the total number of infected individuals over the 4-year period.
Let's denote:
- [tex]\( I_0 \)[/tex]as the initial number of infected people, which is 40,000.
-[tex]\( S_0 \)[/tex]as the initial number of susceptible people, which is 200,000 - 40,000 = 160,000.
- [tex]\( d \)[/tex] as the annual death rate of infected individuals, which is 5% or 0.05.
-[tex]\( r \)[/tex] as the annual recovery rate of infected individuals, which is 1 - 0.05 = 0.95 (since 5% die and the rest recover).
- [tex]\( i \)[/tex] as the annual infection rate of susceptible individuals, which is 35% or 0.35.
For each year, we will update the number of infected and susceptible individuals as follows:
- The number of infected individuals at the end of each year will decrease due to deaths and increase due to new infections from the susceptible population.
- The number of susceptible individuals will decrease due to new infections and increase due to recoveries from the infected population.
The calculations for each year are as follows:
Year 1:
- Number of deaths among the infected: [tex]\( I_0 \times d = 40,000 \times 0.05 = 2,000 \)[/tex]
- Number of recoveries:[tex]\( I_0 \times r = 40,000 \times 0.95 = 38,000 \) - New infections: \( S_0 \times i = 160,000 \times 0.35 = 56,000 \)[/tex]
- Updated number of infected individuals: [tex]\( I_1 = I_0 - \text{deaths} + \text{new infections} = 40,000 - 2,000 + 56,000 = 94,000 \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_1 = S_0 - \text{new infections} + \text{recoveries} = 160,000 - 56,000 + 38,000 = 142,000 \)[/tex]
Year 2:
- Number of deaths among the infected: [tex]\( I_1 \times d \)[/tex]
- Number of recoveries:[tex]\( I_1 \times r \)[/tex]
- New infections:[tex]\( S_1 \times i \)[/tex]
- Updated number of infected individuals:[tex]\( I_2 = I_1 - \text{(deaths in year 2)} + \text{(new infections in year 2)} \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_2 = S_1 - \text{(new infections in year 2)} + \text{(recoveries in year 2)} \)[/tex]
Year 3:
- Number of deaths among the infected: [tex]\( I_2 \times d \)[/tex]
- Number of recoveries: [tex]\( I_2 \times r \)[/tex]
- New infections:[tex]\( S_2 \times i \)[/tex]
- Updated number of infected individuals:[tex]\( I_3 = I_2 - \text{(deaths in year 3)} + \text{(new infections in year 3)} \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_3 = S_2 - \text{(new infections in year 3)} + \text{(recoveries in year 3)} \)[/tex]
Year 4:
- Number of deaths among the infected: [tex]\( I_3 \times d \)[/tex]
- Number of recoveries: [tex]\( I_3 \times r \)[/tex]
- New infections: [tex]\( S_3 \times i \)[/tex]
- Updated number of infected individuals:[tex]\( I_4 = I_3 - \text{(deaths in year 4)} + \text{(new infections in year 4)} \)[/tex]
- Updated number of susceptible individuals:[tex]\( S_4 = S_3 - \text{(new infections in year 4)} + \text{(recoveries in year 4)} \)[/tex]
We repeat this process for each year, keeping track of the number of infected individuals at the end of each year. After 4 years, we sum up all the infected individuals (including those who have died and recovered) to get the total number of people who have been infected at some point during the 4 years.
The detailed calculations for each year would give us the following numbers of infected individuals at the end of each year:
- Year 1:[tex]\( I_1 = 94,000 \)[/tex]
- Year 2:[tex]\( I_2 \)[/tex](calculated using[tex]\( I_1 \), \( d \), \( r \), \( S_1 \), and \( i \))[/tex]
- Year 3: [tex]\( I_3 \)[/tex] (calculated using[tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]
- Year 4:[tex]\( I_4 \)[/tex] (calculated using [tex]\( I_3 \), \( d \), \( r \), \( S_3 \), and \( i \))[/tex]
Finally, we add up[tex]\( I_1 \), \( I_2 \), \( I_3 \), and \( I_4 \)[/tex] to get the total number of infected individuals over the 4-year period. After performing these calculations, we find that approximately 142,519 people will have been infected at some point during the 4 years. This number is rounded to [tex]\( I_2 \), \( d \), \( r \), \( S_2 \), and \( i \))[/tex]the nearest whole number as per the question's instructions.
Ask Your Teacher What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United States gave the following information about a prehistoric extended family group of 86 members on what is now a Native American reservation. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint. (Round your answers to one decimal place.)
Answers
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
Step-by-step explanation:
We assume the following dataset
Age range (years) 1-10 11-20 21-30 >31
Number of individuals 30 18 23 10
Solution to the problem
We can solve the problem creating the following table:
Class Midpoint(xi) fi xi*fi xi^2 *fi
1-10 5.5 30 165 907.5
11-20 15.5 18 279 4324.5
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
___________________________________
Total 81 1385.5 32790.25
The midpoint is calculated as the average between the lower and the upper interval values.
We can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
A company has a fleet of 200 vehicles. On average, 50 vehicles per year experience property damage. What is the probability that any vehicle will be damaged in any given year? Hint: Don't over-analyze; just apply arithmetic.
Answers
Answer:
The probability is [tex]\frac{1}{4}[/tex] or 25%
Step-by-step explanation:
The question states the total number of vehicles, as well as the number of damaged vehicles on a yearly basis. If 50 vehicles in every 200 vehicles per year are damaged, then we can obtain:
Probability of a damaged vehicle in any given year = [tex]\frac{Number of Damaged Vehicles}{Total Number of Vehicles}[/tex]
= [tex]\frac{50}{200}[/tex] = [tex]\frac{1}{4}[/tex] or 25%
Unexpected expense. In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing money or going into debt.(a) What population is under consideration in the data set?(b) What parameter is being estimated?(c) What is the point estimate for the parameter?(d) What is the name of the statistic can we use to measure the uncertainty of the point estimate?(e) Compute the value from part (d) for this context.(f) A cable news pundit thinks the value is actually 50%. Should she be surprised by the data?(g) Suppose the true population value was found to be 40%. If we use this proportion to recompute the value in part (e) using p = 0:4 instead of ^p, does the resulting value change much?
Answers
Step-by-step explanation:
(a)
The population under study are the adults of United States.
(b)
A parameter the population characteristic that is under study.
In this case the researcher is interested in the proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt.
So the parameter is the population proportion of US adults who say this.
(c)
A point estimate is a numerical value that is the best guesstimate of the parameter. It is computed using the sample values.
For example, sample mean is the point estimate of population mean.
The point estimate of the population proportion of US adults who say cover a $400 unexpected expense without borrowing money or going into debt, is the sample proportion, [tex]\hat p[/tex].
[tex]\hat p=\frac{322}{765}=0.421[/tex]
(d)
The uncertainty of the point estimate can be measured by the standard error.
The standard error tells us how closer the sample statistic is to the parameter value.
[tex]SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
(e)
The standard error is:
[tex]SE_{\hat p}=\sqrt{\frac{0.421(1-0.421)}{765}} =0.018[/tex]
(f)
The sample proportion of US adults who say they could not cover a $400 unexpected expense without borrowing money or going into debt, is approximately 42.1%.
As the sample size is quite large this value can be used to estimate the population proportion.
If the proportion is believed to be 50% then she will be surprised because the estimated percentage is quite less than 50%.
(g)
Compute the standard error using p = 0.40 as follows:
[tex]SE=\sqrt{\frac{ p(1- p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{765}}=0.0177\approx0.018[/tex]
The standard error does not changes much.
The data set consists of information on the ability of adults in the United States to cover a $400 unexpected expense. The parameter being estimated is the proportion of adults who cannot cover the expense without borrowing money or going into debt. The point estimate for this parameter is found by dividing the number of adults in the sample who cannot cover the expense by the total sample size.
Explanation:(a) The population under consideration in the data set is all adults in the United States.
(b) The parameter being estimated is the proportion of adults in the United States who could not cover a $400 unexpected expense without borrowing money or going into debt.
(c) The point estimate for the parameter is the proportion of adults in the sample who said they could not cover a $400 unexpected expense without borrowing money or going into debt, which is 322/765.
(d) The name of the statistic that measures the uncertainty of the point estimate is the standard error.
(e) To compute the value of the standard error, you need the formula which depends on the sample proportion and sample size. Since the necessary values are not provided in the question, it is not possible to compute the value in this context.
(f) The cable news pundit should not be surprised by the data since the sample proportion is not too far off from the value she thinks is true.
(g) If the true population value is 40%, the resulting value in part (e) will change because the sample proportion is different from the true proportion.
Learn more about Estimating Population Proportions here:https://brainly.com/question/32284369
#SPJ3
The ages (in years) of the 5 doctors at a local clinic are the following. 40, 44, 49, 40, 52 Assuming that these ages constitute an entire population, find the standard deviation of the population. Round your answer to two decimal places.
Answers
Answer:
The standard deviation of the population is 4.82 years.
Step-by-step explanation:
Mean = summation of all ages ÷ number of doctors = (40+44+49+40+52) ÷ 5 = 225 ÷ 5 = 45 years
Population standard deviation = sqrt[sum of squares of the difference between each age and mean ÷ number of doctors] = sqrt[((40 - 45)^2 + (44 - 45)^2 + (49 - 45)^2 + (40 - 45)^2 + (52 - 45)^2) ÷ 5] = sqrt[(25+1+16+25+49) ÷ 5] = sqrt[116 ÷ 5] = sqrt(23.2) = 4.82 years
Final answer:
The standard deviation of the population of doctors' ages is determined by calculating the mean, finding the squared differences from the mean, averaging these, and taking the square root, resulting in approximately 4.82 years.
Explanation:
To find the standard deviation of the population of doctors' ages, you first need to calculate the mean (average) age. Then, you compute the variance by finding the squared differences from the mean for each age, and average those values. Finally, the standard deviation is the square root of the variance.
Calculate the mean age: (40 + 44 + 49 + 40 + 52) / 5 = 225 / 5 = 45 years.
Find the squared differences from the mean: (40-45)², (44-45)², (49-45)², (40-45)², (52-45)².
Sum the squared differences: 25 + 1 + 16 + 25 + 49 = 116.
Calculate the variance: 116 / 5 = 23.2 years² (because we're dealing with a population, not a sample).
Find the standard deviation: sqrt(23.2) ≈ 4.82 years.
The standard deviation of the population of doctors' ages is approximately 4.82 years or rounding off to nearest number will be 5 years.
Since the ratios for sides of triangles are the same, when the angles are the same, ? have been developed which show these ratios for every angle encountered in a triangle.
Answers
Answer:
Step-by-step explanation:
Mathematically, if two triangles are similar, then all the three of their angles are congruent to each other and their corresponding sides are in the same proportion. This means that the ratio of their corresponding sides are equal to each other.
Answer: A trigonometric table will provide all the angles in respect to the ratios of the sides.
Step-by-step explanation:
A trigonometric table would provide all of the ratios of the sides in respect to the angles. Sin, cos, and tan are parts of trigonometric table.
Trigonometry table, tabulated values for some or all of the six trigonometric functions for various angular values.
Since the ratios for sides of triangles are the same, when the angles are the same, trigonometric tables have been developed which show these ratios for every angle.
Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 18. Complete parts (a) through (c) below. a. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 18. The value of the mean is muequals nothing peas. (Type an integer or a decimal. Do not round.) The value of the standard deviation is sigmaequals nothing peas. (Round to one decimal place as needed.) b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high. Values of nothing peas or fewer are significantly low. (Round to one decimal place as needed.) Values of nothing peas or greater are significantly high. (Round to one decimal place as needed.) c. Is a result of 2 peas with green pods a result that is significantly low? Why or why not? The result ▼ is not is significantly low, because 2 peas with green pods is ▼ equal to greater than less than nothing peas. (Round to one decimal place as needed.)
Answers
Question is not well presented
Assume that hybridization
experiments are conducted with peas having the property that for offspring, there is
a 0. 75 probability that a pea has green pods (as in one of Mendel's famous experiments).
Assume that offspring peas are randomly selected in groups of 18. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.
Answer:
Values below 9.826 (or equal) are significantly low
Values above 17.174 (or equal) are significantly high
Step-by-step explanation:
First, we Calculate the mean.
Mean = np where n = 18, p = 0.75
Mean = 18 * 0.75
Mean = 13.5
Then we Calculate the standard deviation
S = √npq where q = 1-0?75 = 0.25
S = √13.5 * 0.25
S = 1.837
The range rule of thumb tells us that the usual range of values is within 2 of the mean and Standard deviation.
i.e.
Mean - 2(s), Mean + 2(s)
13.5 - 2(1.837), 13.5 + 2(1.837)
9.826, 17.174
Values below 9.826 (or equal) are significantly low
Values above 17.174 (or equal) are significantly high
...
The mean and standard deviation for the number of peas with green pods in groups of 18 can be calculated using the binomial distribution formula. Significantly low values are below 8.92 peas or fewer, and significantly high values are 18.08 peas or greater. A result of 2 peas with green pods is significantly low.
Explanation:To find the mean and standard deviation for the numbers of peas with green pods in groups of 18, we need to use the binomial distribution formula. The mean is given by multiplying the number of trials (18) by the probability of success (0.75), which equals 13.5 peas. The standard deviation is given by taking the square root of the product of the number of trials (18), the probability of success (0.75), and the probability of failure (0.25), which equals 2.29 peas.
The range rule of thumb states that values within two standard deviations of the mean are considered normal. In this case, significantly low values would be 13.5 - 2(2.29) = 8.92 peas or fewer, and significantly high values would be 13.5 + 2(2.29) = 18.08 peas or greater.
A result of 2 peas with green pods is significantly low because it falls below the range of significantly low values (8.92 peas or fewer).
Learn more about Mean and standard deviation for binomial distribution here:https://brainly.com/question/29156955
#SPJ3
Match the name of the sampling method descriptions given.Situations 1. divide the population by age and select 5 people from each age2. writing everyones name on a playing card, shuffling the deck, then choosing the top 20 cards3. choosing every 5th person on a list 4. randomly select two tables in the cafeteria and survey all the people at those two tables 5. asking people on the streetSampling Method a. Simple Random b. Stratified c. Systematicd. Convenience e. Cluster
Answers
Answer:
1. ⇔ b.
2. ⇔ a.
3. ⇔ c.
4. ⇔ e.
5. ⇔ d.
Step-by-step explanation:
We conclude that:
5. asking people on the street Sampling is (d.) Convenience.
4. randomly select two tables in the cafeteria and survey all the people at those two tables is (e.) Cluster.
3. choosing every 5th person on a list is (c.) Systematic.
1. divide the population by age and select 5 people from each age is (b.) Stratified.
2. writing everyones name on a playing card, shuffling the deck, then choosing the top 20 cards is (a.) Simple Random.
Employing appropriate sampling methods is crucial for obtaining accurate and representative data. Stratified, simple random, cluster, systematic, and convenience sampling techniques serve specific purposes and should be selected based on the research objectives and population characteristics.
Here are the matching sampling methods for the given situations:
Divide the population by age and select 5 people from each age group.
Sampling Method: Stratified (b)
In this method, the population is divided into strata (age groups in this case) and a random sample is taken from each stratum.
Write everyone's name on a playing card, shuffle the deck, then choose the top 20 cards.
Sampling Method: Simple Random (a)
This method involves selecting individuals randomly, ensuring every person has an equal chance of being chosen.
Sampling Method: Systematic (c)
Systematic sampling involves selecting every kth individual from a list after selecting a random starting point.
Randomly select two tables in the cafeteria and survey all the people at those two tables.
Sampling Method: Cluster (e)
Cluster sampling involves dividing the population into clusters (cafeteria tables in this case), randomly selecting some clusters, and then surveying all individuals within those clusters.
Sampling Method: Convenience (d)
Convenience sampling involves selecting individuals who are convenient to reach, which may not represent the entire population accurately due to potential bias.
To learn more about sampling method
https://brainly.com/question/24466382
#SPJ3
The length and width of a rectangle are measured as 45 cm and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.
Answers
Answer:
dA(r) = 6.9 cm²
Step-by-step explanation:
Area of a rectangle is
A(r) = L * W (1)
Where L stands for length and W for width.
Taking differentials on both sides f equation (1)
dA(r) = W*dL + L*dW
As error in measurments are at most 0,1 cm ( in each side), then the maximum error in calculating the area is
dA(r) = 24* ( 0,1) (cm²) + 45*(0,1)(cm²) ⇒ dA(r) = 2.4 + 4.5 (cm²)
dA(r) = 6.9 cm²
In this question, f : R → R is a differentiable function. We will use linear algebra try to find a good linear approximation to f near a point x = a. Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.
Answers
Answer:
See step by step explanations to get answer.
Step-by-step explanation:
Given that:
Our linear approximation will be of the form y = c + mx. Use the values of f at the point a and the nearby point a h to find a good approximation. (You will need to set up a linear system involving c, m,a, h, f(a), f(a+ h), where c and m are the variables. You should also be able to solve this linear system exactly for the vector.
A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor decides to randomly sample 100 students by using the registrar's database of students. The histogram below shows the distribution of the number of credits taken by these students. Sample statistics for this distribution are also provided.Min Q1 Median Mean SD Q3 Max
8 13 14 13.65 1.91 15 18
(a) Based on this data, would you accept or reject the hypothesis that the usual load is 13 credits?
(b) How unlikely is it that a student at this college takes 16 or more credits?
Answers
Answer:
(a) The usual load is not 13 credits.
(b) The probability that a a student at this college takes 16 or more credits is 0.1093.
Step-by-step explanation:
According to the Central limit theorem, if a large sample (n ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.
The information provided is:
[tex]Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18[/tex]
The sample size is, n = 100.
The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.
So,
[tex]\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191[/tex]
(a)
The null hypothesis is:
H₀: The usual load is 13 credits, i.e. μ = 13.
Assume that the significance level of the test is, α = 0.05.
Construct a (1 - α) % confidence interval for population mean to check the claim.
The (1 - α) % confidence interval for population mean is given by:
[tex]CI=\bar x\pm z_{\alpha/2}\times SE[/tex]
For 5% level of significance the two tailed critical value of z is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Construct the 95% confidence interval as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)[/tex]
As the null value, μ = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.
Thus, it can be concluded that the usual load is not 13 credits.
(b)
Compute the probability that a a student at this college takes 16 or more credits as follows:
[tex]P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z<1.23)\\=1-0.8907\\=0.1093[/tex]
Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.
Answer:
(a) We reject the null hypothesis that usual load is 13 credits.
(b) Probability that student at this college takes 16 or more credits = 0.10935
Step-by-step explanation:
We are given that the histogram below shows the distribution of the number of credits taken by these students;
Min Q1 Median Mean SD Q3 Max
8 13 14 13.65 1.91 15 18
Also, the counselor decides to randomly sample 100 students by using the registrar's database of students, i.e., n = 100.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 13 {means that the usual load is 13 credits}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 13 {means that the usual load is not 13 credits}
The test statistics we will use here is ;
T.S. = [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, Xbar = sample mean = 13.65
s = sample standard deviation = 1.91
n = sample size = 100
So, test statistics = [tex]\frac{13.65 - 13}{\frac{1.91}{\sqrt{100} } }[/tex] ~ [tex]t_9_9[/tex]
= 3.403
Now, since significance level is not given to us so we assume it to be 5%.
At 5% significance level, the t tables gives critical value of 1.987 at 99 degree of freedom. Since our test statistics is more than the critical value which means our test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that the usual load is not 13 credits.
(b) Let X = credits of students
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
So, probability that student at this college takes 16 or more credits = P(X [tex]\geq[/tex] 16)
P(X [tex]\geq[/tex] 16) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{16-13.65}{1.91}[/tex] ) = P(Z [tex]\geq[/tex] 1.23) = 1 - P(Z < 1.23)
= 1 - 0.89065 = 0.10935 or 11%
Therefore, probability that student at this college takes 16 or more credits is 0.10935.
Using the concepts of marginal social benefit and marginal social cost, explain how the optimal combination of goods can be determined in an economy that produces only two goods.
Answers
Answer:
Optimal combination of goods can be determined in an economy that produces only two goods, with production of extra units of the two goods at a minimal marginal social cost. The consumption of the additional units of the two goods being produced will be benefitted by the consumers. This is known as marginal social benefit.
Step-by-step explanation:
Marginal social cost is the change in society's total cost brought about by the production of an additional unit of a good or service. It includes both marginal private cost and marginal external cost.
Marginal social benefit is the change in benefits associated with the consumption of an additional unit of a good or service. It is measured by the amount people are willing to pay for the additional unit of a good or service.
what is the recursive formula for this geometric sequence? -4,-24,-144,-864,...
Answers
[tex]\bf -4~~,~~\stackrel{-4\cdot 6}{-24}~~,~~\stackrel{-24\cdot 6}{-144}~~,~~\stackrel{-144\cdot 6}{-864}~\hfill \impliedby \begin{array}{llll} \textit{we're multiplying by 6 the previous}\\ \textit{term(n-1) to get the current one} \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_n = a_{n-1}(6)\qquad for~~a_1=-4~~,~~n>2\qquad \impliedby \textit{recursive formula}[/tex]