Answer:
a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.
b) 0.0064% probability that all of the selected adults believe in reincarnation.
c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.
d) Since [tex]P(X \geq 5) < 0.05[/tex], 5 is a significantly high number of adults who believe in reincarnation in this sample.
Step-by-step explanation:
For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [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]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 6, p = 0.2[/tex]
a. What is the probability that exactly five of the selected adults believe in reincarnation?
This is P(X = 5).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015[/tex]
There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.
b. What is the probability that all of the selected adults believe in reincarnation?
This is P(X = 6).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064[/tex]
There is a 0.0064% probability that all of the selected adults believe in reincarnation.
c. What is the probability that at least five of the selected adults believe in reincarnation?
This is
[tex]P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564[/tex]
There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.
d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?
5 is significantly high if [tex]P(X \geq 5) < 0.05[/tex]
We have that
[tex]P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05[/tex]
Since [tex]P(X \geq 5) < 0.05[/tex], 5 is a significantly high number of adults who believe in reincarnation in this sample.
a. The probability that exactly five of the selected adults believe in reincarnation is approximately 0.00256.
b. The probability that all of the selected adults believe in reincarnation is approximately 0.000064.
c. The probability that at least five of the selected adults believe in reincarnation is approximately 0.002624.
d. To determine if five is significantly high, we need a significance level for comparison, which isn't provided in the question.
To solve this problem, we can use the binomial probability formula, where "n" is the number of trials, "p" is the probability of success (believing in reincarnation in this case), and "x" is the number of successes.
a. The probability that exactly five of the selected adults believe in reincarnation is calculated as follows:
P(X = 5) = C(6, 5) * (0.20)^5 * (0.80)^(6-5),
where C(6, 5) is the number of ways to choose 5 out of 6 adults, which equals 6.
P(X = 5) = 6 * (0.20)^5 * (0.80)^1 ≈ 0.00256
b. The probability that all of the selected adults believe in reincarnation is:
P(X = 6) = (0.20)^6 ≈ 0.000064
c. The probability that at least five of the selected adults believe in reincarnation is the sum of the probabilities from parts (a) and (b):
P(X ≥ 5) = P(X = 5) + P(X = 6) ≈ 0.00256 + 0.000064 ≈ 0.002624
d. To determine if five is a significantly high number who believe in reincarnation, we can compare the probability of getting at least five believers (from part c) to a significance level. If this probability is less than the significance level, it would be considered significant. The significance level would depend on the context and what is considered "significant" in the specific analysis.
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complete question should be :
See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly selected, and find the indicated probability. a. What is the probability that exactly five of the selected adults believe in reincarnation? b. What is the probability that all of the selected adults believe in reincarnation? c. What is the probability that at least five of the selected adults believe in reincarnation? d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation .
A bag of peanuts in their shells contains 181 peanuts. 65 of the shells contain one peanut, 111 of the shells contain two peanuts, and the rest contain three peanuts. Of the shells with one peanut, 16 of them are cracked. Of the shells with two peanuts, 30 of them are cracked. None of the shells with three peanuts are cracked. One peanut shell is randomly selected from the bag.
A. What is the probability the shell is cracked?B. What is the probability the shell is not cracked and it contains two peanuts?C. What is the probability the shell is not cracked or it contains two peanuts?D. What is the probability the shell is not cracked given that it contains two peanuts?
Answer:
Step-by-step explanation:
Hello!
You have a bag with 181 peanuts in it.
65 of the shells contain one peanut. ⇒ 16 are cracked and 49 are complete.
111 of the shells contain two peanuts. ⇒ 30 are cracked 81 are complete.
5 of the shells have three peanuts and none of them is cracked.
A. What is the probability the shell is cracked?
To calculate the probability of the shell bein cracked you have to add all possible cases in which it is cracked and divide it by the total of peanuts in the bag:
[tex]P(Cracked)= \frac{(16+30)}{181}= 0.25[/tex]
B. What is the probability the shell is not cracked (Cr) and it contains two peanuts(2p)?
This probability is an intersection P(Cr∩2p), to calculate it you have to divide the total of peanuts that fulfill these characteristics and divide it by the total number of peanuts in the bag.
[tex]P(Crn2p)= \frac{30}{181}= 0.17[/tex]
C. What is the probability the shell is not cracked (Cr') or it contains two peanuts(2p)?
This probability is the union between two events, the event "cracked" and the event "two peanuts", symbolically: P(Cr'∪2p), these two events are not mutually exclusive, this means that they can happen at the same time, you have to apply the following formula to calculate it:
[tex]P(Cr'u2p)= P(Cr') + P(2p) - P(Cr'n2p)[/tex]
Since these events aren't mutually exclusive, some of them fulfill both categories, this means they are counted when you calculate the probability of "not cracked" and again when you calculate the probability of "2 peanuts" therefore you need to subtract their intersection to obtain the correct probability.
[tex]P(Cr'u2p)= \frac{(49+81+5)}{181} + (\frac{111}{181} ) - (\frac{81}{181} )= 0.76+0.61-0.45=0.92[/tex]
D. What is the probability the shell is not cracked given that it contains two peanuts?
This is a conditional probability, this means that both events are dependant and the occurrence of "2p" modifies the probability of the peanut shell to be "Cr'", symbolically: P(Cr'/2p) and you calculate it using the following formula:
[tex]P(Cr'/2p)= \frac{P(Cr'n2p)}{P(2p)}= \frac{0.45}{0.61}= 0.74[/tex]
I hope it helps!
A) The probability the shell is cracked is 25.4%.
B) The probability the shell is not cracked and it contains two peanuts is 44.75%.
C) The probability the shell is not cracked or it contains two peanuts is 91.16%.
D) The probability the shell is not cracked given that it contains two peanuts is 72.97%.
Since a bag of peanuts in their shells contains 181 peanuts. 65 of the shells contain one peanut, 111 of the shells contain two peanuts, and the rest contain three peanuts, and, of the shells with one peanut, 16 of them are cracked, and of the shells with two peanuts, 30 of them are cracked, while none of the shells with three peanuts are cracked, and one peanut shell is randomly selected from the bag, to determine A) what is the probability the shell is cracked; B) what is the probability the shell is not cracked and it contains two peanuts; C) what is the probability the shell is not cracked or it contains two peanuts; and D) what is the probability the shell is not cracked given that it contains two peanuts; the following calculations must be performed:
A)
181 = 10016 + 30 = X46 x 100/181 = X25.4 = XB)
181 = 100111 - 30 = X81 x 100 / 181 = X44.75 = XC)
181 = 100111 + 5 + 49 = X165 x 100 / 181 = X91.16 = XD)
111 = 100111 - 30 = X81 x 100 / 111 = X72.97 = XLearn more in https://brainly.com/question/18373730
An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58 type Q batteries. The mean voltage is measured as 8.54 for the type K batteries with a standard deviation of 0.225, and the mean voltage is 8.69 for type Q batteries with a standard deviation of 0.725. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1 be the true mean voltage for type K batteries and μ2 be the true mean voltage for type Q batteries.Use a 0.1 level of significance.
Answer:
Hypothesis Test states that we will accept null hypothesis.
Step-by-step explanation:
We are given that an engineer is comparing voltages for two types of batteries (K and Q).
where, [tex]\mu_1[/tex] = true mean voltage for type K batteries.
[tex]\mu_2[/tex] = true mean voltage for type Q batteries.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 = \mu_2[/tex] {mean voltage for these two types of
batteries is same}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu_1 \neq \mu_2[/tex] {mean voltage for these two types of
batteries is different]
The test statistics we use here will be :
[tex]\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] follows [tex]t_n__1 + n_2 -2[/tex]
where, [tex]X_1bar[/tex] = 8.54 and [tex]X_2bar[/tex] = 8.69
[tex]s_1[/tex] = 0.225 and [tex]s_2[/tex] = 0.725
[tex]n_1[/tex] = 37 and [tex]n_2[/tex] = 58
[tex]s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2} }{37+58-2} }[/tex] = 0.585 Here, we use t test statistics because we know nothing about population standard deviations.
Test statistics = [tex]\frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58} } }[/tex] follows [tex]t_9_3[/tex]
= -1.219
At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that mean voltage for these two types of batteries is same.
The toasters produced by a company have a normally distributed life span with a mean of 5.8 years and a standard deviation of 0.9 years, what warranty should be provided so that the company is replacing at most 10% of their toasters sold? a. 4.3 years b. 5.9 years c. 4.5 years d. 4.6 years
Answer:
d. 4.6 years
Step-by-step explanation:
Problems of normally distributed samples can be 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.
In this problem, we have that:
[tex]\mu = 5.8, \sigma = 0.9[/tex]
What warranty should be provided so that the company is replacing at most 10% of their toasters sold?
Only those on the 10th percentile or lower will be replaced.
So the warranty is the value of X when Z has a pvalue of 0.10.
So it is X when [tex]Z = -1.28[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 5.8}{0.9}[/tex]
[tex]X - 5.8 = -1.28*0.9[/tex]
[tex]X = 4.6[/tex]
So the correct answer is:
d. 4.6 years
Final answer:
To find the warranty length, we need to determine the value that corresponds to the 10th percentile of the normally distributed life span. By solving the equation using the z-score formula, we find that the warranty should be provided for at least 4.6 years.
Explanation:
In order to determine the warranty that should be provided to the toasters, we need to find the value that corresponds to the 10th percentile of the normally distributed life span. To do this, we can use a standard normal distribution table or a calculator. The z-score for the 10th percentile is approximately -1.28. Using the formula z = (x - mean) / standard deviation, we can solve for x to find the warranty length.
-1.28 = (x - 5.8) / 0.9
x - 5.8 = -1.28 × 0.9
x - 5.8 = -1.152
x = 5.8 - 1.152
x = 4.648
Therefore, the warranty should be provided for at least 4.648 years, which is approximately 4.6 years. The correct answer is option d. 4.6 years.
A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 380.(a) Find an expression for the number of bacteria after t hours.P(t)
Answer:
P(t) = 100(3.8^t)
Step-by-step explanation:
You want a function P(t) that describes the exponential population growth of a bacteria culture from 100 cells to 380 in one hour, where t is in hours.
Exponential functionThe function can be written in the form ...
population = (initial population) × (growth factor)^(t/(growth period))
Here, the initial population is 100, and the growth factor in a period of 1 hour is 380/100 = 3.8. Since we want t in hours, this is ...
population = 100 × 3.8^(t/1)
P(t) = 100(3.8^t)
Horalco's garden is shown below. He needs 1 1/3 scoops of fertilizer for each square foot of the garden. How many scoops of fertilizer does Horaclo need for the entire garden? (Garden is 10 1/2 ft by 5 1/2 ft)
Answer: he would need 77 scoops of fertilizer.
Step-by-step explanation:
The measure of Horalco's garden is
10 1/2 ft by 5 1/2 ft. Converting to improper fraction, it becomes 21/2 ft by 11/2 ft. The garden is rectangular in shape and the formula for determining the area of a rectangle is expressed as
Area = length × width.
Therefore, area of the garden is
21/2 × 11/2 = 231/4 ft²
He needs 1 1/3 scoops of fertilizer for each square foot of the garden. Converting to improper fraction, it becomes 4/3 scoops. Therefore, the number of scoops of fertilizer that Horaclo need for the entire garden is
4/3 × 231/4 = 77 scoops of fertilizer.
the average monthly precipitation for honolulu, hi for october, november, and december is 3.11 in. If 2.98. falls in October & 3.05 in. falls in November, how many inches must fall in December so that the average monthly precipitation for these months exceeds 3.11 in
To find out how many inches must fall in December so that the average monthly precipitation for October, November, and December exceeds 3.11 inches, subtract the precipitation from October and November from the total precipitation. Therefore, 3.3 inches must fall in December.
To find out how many inches must fall in December so that the average monthly precipitation for October, November, and December exceeds 3.11 inches, we can use the formula for calculating average:
Average = (Total precipitation) / (Number of months)
Let's solve for the total precipitation:
October precipitation: 2.98 inchesNovember precipitation: 3.05 inchesAverage precipitation = 3.11 inchesTotal precipitation = Average precipitation * Number of months
Total precipitation = 3.11 inches * 3 months = 9.33 inches
To find out how many inches must fall in December, we subtract the precipitation from October and November from the total precipitation:
December precipitation = Total precipitation - (October precipitation + November precipitation)
December precipitation = 9.33 inches - (2.98 inches + 3.05 inches) = 3.3 inches
Therefore, in order for the average monthly precipitation for these months to exceed 3.11 inches,
3.3 inches
must fall in December.
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According to records in a large hospital, the birth weights of newborns have a symmetric and bell-shaped relative frequency distribution with mean 6.8 pounds and standard deviation 0.5 Approximately what proportion of babies are born with birth weight under 6.3 pounds?
Answer:
15.9% of babies are born with birth weight under 6.3 pounds.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 6.8 pounds
Standard Deviation, σ = 0.5
We are given that the distribution of birth weights is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(birth weight under 6.3 pounds)
P(x < 6.3)
[tex]P( x < 6.3) = P( z < \displaystyle\frac{6.3 - 6.8}{0.5}) = P(z < -1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < -1) = 0.159 = 15.9\%[/tex]
15.9% of babies are born with birth weight under 6.3 pounds.
To find the proportion of babies born with a weight less than 6.3 pounds, we calculate the Z-score which results in -1. This Z-score corresponds to about 16% of the population in a standard normal distribution. Therefore, roughly 16% of babies are born weighing less than 6.3 pounds.
Explanation:To solve this problem, you can use the properties of a normal distribution. In a normal distribution, scores fall within one standard deviation of the mean approximately 68% of the time, within two standard deviations about 95% of the time, and within three standard deviations about 99.7% of the time.
In this scenario, we would find the Z-score, a measure that describes a value's relationship to the mean of a group of values. The formula for the Z-score is (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.
With a mean (μ) = 6.8 pounds, standard deviation (σ) = 0.5 pounds, and X = 6.3 pounds, the calculation for the Z-score would be (6.3-6.8)/0.5 = -1. This means that 6.3 pounds is one standard deviation below the mean. Referring to the standard normal distribution table, a Z-score of -1 corresponds to approximately 16% or 0.16 of the population. Therefore, approximately 16% of babies are born with a weight of less than 6.3 pounds.
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Gloria, Tina, and Kelly went to an office supply store. Gloria bought 5 pencils, 7 markers, and 8 erasers. Her total was $19.00. Tina spent $21.00 buying 7 pencils, 5 markers, and 8 erasers. Kelly bought 4 pencils, 8 markers, and 7 erasers for $17.75. What is the cost of each item?
Answer:
Pencils cost $2
Markers cost $1
Erasers cost $0.25
Step-by-step explanation:
Let pencils be denoted by P, markers by M and erasers by E. The following linear system can be modeled from the data provided:
[tex]5P+7M+8E = 19\\7P+5M+8E=21\\4P+8M+7E =17.75[/tex]
Solving the linear system:
[tex]2P-2M= 2\\P=1+M\\4+4M+8M+7E =17.75\\\\12M+7E = 13.75\\12M+8E=14\\E=0.25\\M=\frac{14-0.25*8}{12}\\ M=1\\P=1+1=2[/tex]
Pencils cost $2
Markers cost $1
Erasers cost $0.25
Answer:
Step-by-step explanation:
The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions:
A) What is the probability of completing the exam in ONE hour or less?
B) what is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes?
C) Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to compete the exam in the allotted time?
Answer:
A) P=2.28%
B) P=28.57%
C) I expect 10 students to be unable to complete the exam in the alloted time.
Step-by-step explanation:
In order to solve this problem, we will need to find the respective z-scores. The z-scores are found by using the following formula:
[tex]z=\frac{x-\mu}{sigma}[/tex]
Where:
z= z-score
x= the value to normalize
[tex]\mu = mean[/tex]
[tex]\sigma[/tex]= standard deviation
The z-score will help us find the area below the normal distribution curve, so in order to solve this problem we need to shade the area we need to find. (See attached picture)
A) First, we find the z-score for 60 minutes, so we get:
[tex]z=\frac{60-80}{10}=-2[/tex]
So now we look for the z-score on our normal distribution table. Be careful with the table you are using since some tables will find areas other than the area between the mean and the desired data. The table I used finds the area between the mean and the value to normalize.
so:
A=0.4772 for a z-score of -2
since we want to find the number of students that take less than 60 minutes, we subtract that decimal number from 0.5, so we get:
0.5-0.4772=0.0228
therefore the probability that a student finishes the exam in less than 60 minutes is:
P=2.28%
B) For this part of the problem, we find the z-score again, but this time for a time of 75 minutes:
[tex]z=\frac{75-80}{10}=-0.5[/tex]
and again we look for this z-score on the table so we get:
A=0.1915 for a z-score of -0.5
Now that we got this area we subtract it from the area we found for the 60 minutes, so we get:
0.4772-0.1915=0.2857
so there is a probability of P=28.57% of chances that the students will finish the test between 60 and 75 minutes.
C) Finally we find the z-score for a time of 90 minutes, so we get:
[tex]z=\frac{90-80}{10}=1[/tex]
We look for this z-score on our table and we get that:
A=0.3413
since we need to find how many students will take longer than 90 minutes to finish the test, we subtract that number we just got from 0.5 so we get:
0.5-0.3413=0.1586
this means there is a 15.86% of probabilities a student will take longer than 90 minutes. Now, since we need to find how many of the 60 students will take longer than the 90 available minutes, then we need to multiply the total amount of students by the percentage we previously found, so we get:
60*0.1586=9.516
so approximately 10 Students will be unavailable to complete the exam in the allotted time.
Answer:
Part A:
Probability is P(z<-2)=1-0.9772=0.0228
Part B:
P(-2<z<-0.5)=0.2857
Part C:
Number of students unable to complete the exam=60-50=10 students
Step-by-step explanation:
Part A:
Mean=μ=80 min
Standard Deviation=σ=10 min
Formula:
[tex]z=\frac{\bar x- \mu}{\sigma}[/tex]
where:
[tex]\bar x=60\ min[/tex]
[tex]z=\frac{60-80}{10}\\ z=-2[/tex]
Probability is P(z<-2)
From the probability distribution tables (Cumulative Standardized normal distribution table)
Probability is P(z<-2)=1-0.9772=0.0228
Part B:
For 75 min:
[tex]z=\frac{75-80}{10}\\ z=-0.5[/tex]
For [tex]\bar x=60\ min[/tex]
[tex]z=\frac{60-80}{10}\\ z=-2[/tex]
From the probability distribution tables (Cumulative Standardized normal distribution table)
P(-2<z<-0.5)=P(z<-0.5)-P(z<-2)
P(-2<z<-0.5)=(1-0.6915)-(1-0.9772)
P(-2<z<-0.5)=0.2857
Part C:
[tex]\bar x=90\ min[/tex]
[tex]z=\frac{90-80}{10}\\ z=1[/tex]
From the probability distribution tables (Cumulative Standardized normal distribution table)
Probability is P(z<1)=0.8413
Number of students=0.8413*60
Number of students to complete the exam=50.478≅50
Number of students unable to complete the exam=60-50=10 students
For each of the initial conditions below, determine the largest interval a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.
y(-7) = -5.5
y(-0.5) = 6.4.
y(0) = 0.
y(5.5) = -3.14.
y(10) = 2.6.
The largest intervals on which the existence and uniqueness theorem guarantees a unique solution for the given initial conditions.
Explanation:The interval a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution depends on the initial conditions. For the given initial conditions:
y(-7) = -5.5: The largest interval is from t = -7 to t = -0.5y(-0.5) = 6.4: The largest interval is from t = -0.5 to t = 0y(0) = 0: The largest interval is from t = 0 to t = 5.5y(5.5) = -3.14: The largest interval is from t = 5.5 to t = 10y(10) = 2.6: The largest interval is from t = 10 to t = 74The distribution of a sample of the outside diameters of PVC pipes approximates a normal distribution. The mean is 14.0 inches, and the standard deviation is 0.1 inches. About 68% of the outside diameters lie between what two amounts? A. 13.9 and 14.1 inches B. 13.0 and 15.0 inches C. 13.8 and 14.2 inches D. 13.5 and 14.5 inches
Answer:
A. 13.9 and 14.1 inches
See explanation below.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the outside diameters of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(14,0.1)[/tex]
Where [tex]\mu=14[/tex] and [tex]\sigma=0.1[/tex]
If we want the middle 68% of the data we need to have on the tails 16% on each one
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.84[/tex] (a)
[tex]P(X<a)=0.16[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.16 of the area on the left and 0.84 of the area on the right it's z=-0.994. On this case P(Z<-0.994)=0.16 and P(z>-0.994)=0.84
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.16[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.16[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.994<\frac{a-14}{0.1}[/tex]
And if we solve for a we got
[tex]a=14 -0.994*0.1=13.9[/tex]
And since the distribution is symmetrical for the upper limit we can use z = 0.994 and we have:
[tex]z=0.994<\frac{a-14}{0.1}[/tex]
And if we solve for a we got
[tex]a=14 +0.994*0.1=14.1[/tex]
So the correct answer for this case would be:
A. 13.9 and 14.1 inches
Find k. HELP ME PLEASE PLEASE
Answer:
8
Step-by-step explanation:
Sin 30 = k/16
k = 16 x sin 30
k = 16 x (0.5) = 8
Find the approximate area of the shaded region below, consisting of a square with a circle cut out of it. Use 3.14 as an approximation for PI
A. 856 square feet
B. 86 square feet
C. 314 square feet
D. 214 square feet
Answer:
B
Step-by-step explanation:
Since the figure was not supplied, let's focus on that principle.
To calculate it simply subtract the area of the square minus the area of the circle. Given the side of the square 20 ft
1. Square Area
[tex]S= s^{2}\\S=20^{2} \Rightarrow S=400 ft^{2}[/tex]
2.Circle Area
Notice the radius is half the square side, i.e. 10 ft
[tex]S=\pi*R^{2}\\S=3.14*(10)^{2}\\S=314 \:ft^{2}\\[/tex]
Subtracting the area of the square and the area of the circle:
[tex]400-314=86 ft^{2}[/tex]
The area of the shaded is 86 feet².
Thus, option (B) is correct.
Let's assume the side length of the square is "s".
The area of the square is then given by s².
Substitute the side s = 20 into above formula as
Area of square = 20 x 20
= 400 square feet
Now, Area of circle = πr²
= 3.14 x (10)²
= 3.14 x 100
= 314 square feet.
Now, the area of the shaded region can be calculated as:
Area of shaded region = Area of square - Area of circle
= 400 - 314
= 86 feet²
Thus, option (B) is correct.
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Suppose we have a test for a certain disease. If a person has the disease,the test has 95% chance to be positive. If a person is healthy, the test has 5% chanceto give a false positive result. We assume that a person has 10% chance to have thedisease. Given that the test for a person is positive, what is the probability that theperson has the disease?
Answer:
P ( disease | positive ) = 0.68
Step-by-step explanation:
This is a classic question of Bayes' Theorem, which calculates probability in a scenario where one event has already occured.
In this case, we have the following data:
P ( disease ) = 0.10
P (no disease) = 0.90
P ( positive | disease ) = 0.95
P ( positive | no disease) = 0.05
Formula to use:
[tex]P ( disease\ |\ positive ) = \frac{P(disease)\ P(positive\ |\ disease)}{P(disease)\ P(positive\ |\ disease) + P(no\ disease)\ P(positive\ |\ no\ disease)}[/tex]We substitute the values from our data in this formula:
[tex]\frac{(0.10) \times (0.95)}{(0.10)\times(0.95) + (0.90)\times(0.05)} \\\\\frac{0.095}{0.095+0.045}\\\\\frac{0.095}{0.14}\\ \\\\\=0.68[/tex] (Rounded off to two decimal places)
Hence, the probability of a person testing positive once they have the disease is 68%.
Given the value of cos 50° ? 0.6428, enter the sine of a complementary angle. Use an expression relating trigonometric ratios of complementary angles.
sin 40° = 0.6428
Step-by-step explanation:
The complementary ratio of sine is cos and vice-versa.
sin θ = cos (90 - θ)
cos θ = sin (90 - θ)
So cos 50° = sin (90 - 50)°
= sin 40°
sin 40° = 0.6428
This indicates that the complement of cos 50° is sin 40° which is equal to 0.6428.
Suppose that the national average for the math portion of the College Board's SAT is 540. The College Board periodically rescales the test scores such that the standard deviation is approximately 50. Answer the following questions using a bell-shaped distribution and the empirical rule for the math test scores. If required, round your answers to two decimal places. If your answer is negative use "minus sign".
(a) What percentage of students have an SAT math score greater than 615?
(b) What percentage of students have an SAT math score greater than 690?
(c) What percentage of students have an SAT math score between 465 and 540?
(d) What is the z-score for student with an SAT math score of 625?
(e) What is the z-score for a student with an SAT math score of 415?
Answer:
a) 6.68%
b) 0.135%
c) 43.32%
d) 1.7
e) -2.5
Step-by-step explanation:
a)
P(X>615)=P(z>(615-540)/50)
P(X>615)=P(z>1.5)
P(X>615)=P(0<z<∞)-P(0<z<1.5)
P(X>615)=0.5-0.4332
P(X>615)=0.0668
The percentage of students have an SAT math score greater than 615 is 6.68%
b)
P(X>690)=P(z>(690-540)/50)
P(X>690)=P(z>3)
P(X>690)=P(0<z<∞)-P(0<z<3)
P(X>690)=0.5-0.49865
P(X>690)=0.00135
The percentage of students have an SAT math score greater than 690 is 0.135%
c)
P(465<X<540)=P((465-540)/50<z<(540-540)/50))
P(465<X<540)=P(-1.5<z<0)
P(465<X<540)=0.4332
The percentage of students have an SAT math score between 465 and 540 is 43.32%
d)
z=(625-540)/50
z=85/50
z=1.7
The z-score for student with an SAT math score of 625 is 1.7.
e)
z=(415-540)/50
z=-125/50
z=-2.5
The z-score for a student with an SAT math score of 415 is -2.5.
A cube-shaped water tank having 6 ft side lengths is being filled with water. The bottom is solid metal but the sides of the tank are thin glass which can only withstand a maximum force of 200 lb. How high (in ft) can the water reach before the sides shatter?
(Assume a density of water rho = 62.4 lb/ft3.
Round your answer to two decimal places.)
The distance the in which the water will reach before it shatters will be 1.03ft
Data;
Let the integral run from surface (y = 0) to maximum depth (y = d)ftLet the differential depth = dydtThe area of each depth = 6dy ft^2The Force Acting at any DepthUsing integration, the force at any depth y is 62.4y lb/ft^3
[tex]200 = \int\limits^d_0 {62.4y} \, 6dy\\ 200 = \int\limits^d_0 {374.4y} \, dy\\ 200 = \int\limits^6_0 {374.4y} \, dy\\ 200 = [374.4/2 y^2]_0^d\\200 = 187.2(d^2 - 0)\\200 = 187.2d^2\\d^2 = 200/187.2\\d = 1.03ft[/tex]
From the calculations above, the distance in which the water will reach before it shatters is 1.03ft
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A 400-gal tank initially contains 100 gal of brine contain-ing50lbofsalt. Brinecontaining 1lbofsaltpergallon enters the tank at the rate of 5 gal=s, and the well-mixed brine in the tank flows out at the rate of 3 gal=s. How much salt will the tank contain when it is full of brin?
Answer:395.75lb
Step-by-step explanation:see attachment
True or false: A) Any two different points must be collinear. B) Four points can be collinear. C) Three or more points must be collinear.
Answer:
A) True
B) True
C) False
Step-by-step explanation:
Knowing that the collinear points are all those that pass through a line, we have:
A) given two points they form a line, by themselves they are collinear (graph 1)
B) Can be or not can be (graph 2)
C) Can be not must be (graph 3)
We want to see if the given statements are true or false.
We will see that:
a) trueb) truec) false.What are collinear points?Two or more points are collinear if we can draw a line that connects them.
Analyzing the statements:A) Whit that in mind, the first statement is true, 2 points is all we need to draw a line, thus two different points are always collinear, so the first statement is true.
B) For the second statement suppose you have a line already drawn, then you can draw 4 points along the line, if you do that, you will have 4 collinear points, so yes, 4 points can be collinear.
C) For the final statement, again assume you have a line, you used 2 points to draw that line (because two points are always collinear). Now you could have more points outside the line, thus, the set of all the points is not collinear (not all the points are on the same line).
So sets of 3 or more points can be collinear, but not "must" be collinear, so the last statement is false.
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3)A triangle has all integer side lengths and two of those sides have lengths 9 and 16. Consider the altitudes to the three sides. What is the largest possible value of the ratio of any two of those altitudes
Answer: 2
Step-by-step explanation:
16/9=1.7
=Approximately=2
Answer:
2 is the largest possible value.
Step-by-step explanation:
Given Data:
Length of one side = 9
Length of second side = 16
Let AB and BC sides of triangle be of length 9 and 16 respectively as shown in figure attached.
Now, let sides AD and CF be the respective altitudes.
Also, ∠ABC = ∅ (as shown in figure)
If AD and CF are the respective altitudes then,
we have
AD = 9Sin∅ ;
CF = 16Sin∅;
By dividing both sides, we get
AD/CF = 9/16
This equation shows that is independant of angle ∅.
Now, let ∠BAC = α
Now we have, BE = 9 sinα
and FC = AC sinα
By dividing both sides, we get
BE/FC=9/AC
Similarly we also have,
BE/AD = 16/AC
ABC is a triangle as long as length of AC is ithin range of 8 to 24 i.e, 8≤AC≤24 (because sum of any two sides of triangle should be greater than length of third side)
Using these values we get ranges of:
9/24 ≤ BF/FC ≤ 9/8 ; 2/3 ≤ BE/AD ≤ 2
So,
2 is the largest possible value of the ratio of any two of these altitudes.
A survey regarding the communication habits of Americans was conducted. The study randomly selected 700 American citizens and asked them a series of questions about their daily communication methods. Three primary methods of communication were identified as social media (M), SMS text messaging (T), cell phone call (C). The results are summarized below.
480 people responded using M or T
300 people use all three primary forms of comunication
465 people responded using M or C
440 people responded using C or T
405 respondents use at least 2 of the primary forms of comunication
There was nobody who used both M and C but did not use T
210 respondents used only forms of communication that differed from the primary 3
385 people use both C and T
a. Create a well labeled Venn diagram representing the survey.
b. How many people in the survey used only 1 of the three primary forms of communication?
c. What is the probability that a survey respondent used SMS text messaging?
d. What is the probability that a survey respondent used exactly 2 of the 3 primary forms of communication?
e. Given that a survey respondent used at least 1 of the 3 primary forms of communication, what is the probability that they used all 3?
f. Given a respondent used social media and standard calling with a cell phone, what is the probability that they also used SMS text messaging?
Answer:
answers and explanation is shown in the attachment
Step-by-step explanation:
The detailed steps and necessary substitution is as shown in the attached files.
Let X1 and X2 be independent random variables with mean μand variance σ².
Suppose that we have 2 estimators of μ:
θ₁^ = (X1+X2)/2
θ₂^ = (X1+3X2)/4
a) Are both estimators unbiased estimators ofμ?
b) What is the variance of each estimator?
Answer:
a) [tex] E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu[/tex]
So then we conclude that [tex] \hat \theta_1[/tex] is an unbiased estimator of [tex]\mu[/tex]
[tex] E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu[/tex]
So then we conclude that [tex] \hat \theta_2[/tex] is an unbiased estimator of [tex]\mu[/tex]
b) [tex] Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2} [/tex]
[tex] Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8} [/tex]
Step-by-step explanation:
For this case we know that we have two random variables:
[tex] X_1 , X_2[/tex] both with mean [tex]\mu = \mu[/tex] and variance [tex] \sigma^2[/tex]
And we define the following estimators:
[tex] \hat \theta_1 = \frac{X_1 + X_2}{2}[/tex]
[tex] \hat \theta_2 = \frac{X_1 + 3X_2}{4}[/tex]
Part a
In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:
[tex] E(\hat \theta_i) = \mu , i = 1,2 [/tex]
So let's find the expected values for each estimator:
[tex] E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})[/tex]
Using properties of expected value we have this:
[tex] E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu[/tex]
So then we conclude that [tex] \hat \theta_1[/tex] is an unbiased estimator of [tex]\mu[/tex]
For the second estimator we have:
[tex]E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})[/tex]
Using properties of expected value we have this:
[tex] E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu[/tex]
So then we conclude that [tex] \hat \theta_2[/tex] is an unbiased estimator of [tex]\mu[/tex]
Part b
For the variance we need to remember this property: If a is a constant and X a random variable then:
[tex] Var(aX) = a^2 Var(X)[/tex]
For the first estimator we have:
[tex] Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})[/tex]
[tex] Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)] [/tex]
Since both random variables are independent we know that [tex] Cov(X_1, X_2 ) = 0[/tex] so then we have:
[tex] Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2} [/tex]
For the second estimator we have:
[tex] Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})[/tex]
[tex] Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)] [/tex]
Since both random variables are independent we know that [tex] Cov(X_1, X_2 ) = 0[/tex] so then we have:
[tex] Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8} [/tex]
Both θ₁^ and θ₂^ are unbiased estimators of μ. The variance of θ₁^ is σ² / 2, while the variance of θ₂^ is 5σ² / 8.
Let's analyze the provided estimators of the mean (">").
a) Unbiased Estimators
An estimator heta is unbiased if E[θ] = μ. We have:
θ₁^ = (X₁ + X₂) / 2
The expected value of θ₁^ is:
E[θ₁^] = E[(X₁ + X₂) / 2] = (E[X₁] + E[X₂]) / 2 = (μ + μ) / 2 = μ
Thus, θ₁^ is an unbiased estimator of μ.
θ₂^ = (X₁ + 3X₂) / 4
The expected value of θ₂^ is:
E[θ₂^] = E[(X₁ + 3X₂) / 4] = (E[X₁] + 3E[X₂]) / 4 = (μ + 3μ) / 4 = 4μ / 4 = μ
Thus, θ₂^ is also an unbiased estimator of μ.
b) Variance of Each Estimator
The variance of an estimator θ is given by Var(θ). Considering the given variances of X₁ and X₂ (both σ²):
For θ₁^:Var(θ₁^) = Var[(X₁ + X₂) / 2] = (1/2)²Var(X₁) + (1/2)²Var(X₂) = (1/4)σ² + (1/4)σ² = σ² / 2
For θ₂^:Var(θ₂^) = Var[(X₁ + 3X₂) / 4] = (1/4)²Var(X₁) + (3/4)²Var(X₂) = (1/16)σ² + (9/16)σ² = (1/16 + 9/16)σ² = (10/16)σ² = 5σ² / 8
Thus, the variance of θ₁^ is σ² / 2 and the variance of θ₂^ is 5σ² / 8.
"Assume the average price of a laptop computer is $775 with a standard deviation of $75. The following data represent the prices of a sample of laptops at an electronics store. Calculate the z-score for each of the following prices"
a. $699
b. $949
c. $625
d. $849
e. $999
Answer:
a) -1.013
b) 2.32
c) -2
d) 0.9867
e) 2.9867
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $775
Standard Deviation, σ = $75
We are given that the distribution of average price of a laptop is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find z-score for corresponding
a. $699
[tex]z = \displaystyle\frac{699 - 775}{75} = -1.013[/tex]
b. $949
[tex]z = \displaystyle\frac{949 - 775}{75} = 2.32[/tex]
c. $625
[tex]z = \displaystyle\frac{625 - 775}{75} = -2[/tex]
d. $849
[tex]z = \displaystyle\frac{849 - 775}{75} = 0.9867[/tex]
e. $999
[tex]z = \displaystyle\frac{999 - 775}{75} = 2.9867[/tex]
a. The z-score for $699 is -1.20.
b. The z-score for $949 is 2.13.
c. The z-score for $625 is -2.33.
d. The z-score for $849 is 0.67.
e. The z-score for $999 is 3.13.
To calculate the z-score for each price, you can use the formula:
[tex]\[Z = \frac{X - \mu}{\sigma}\][/tex]
Where:
- [tex]\(Z\)[/tex] is the z-score.
- [tex]\(X\)[/tex] is the value you want to calculate the z-score for (the given price).
- [tex]\(\mu\)[/tex] is the mean (average) price of laptops, which is $775 in this case.
- [tex]\(\sigma\)[/tex]is the standard deviation of laptop prices, which is $75.
Now, let's calculate the z-scores for the given prices:
a. For $699: [tex]\(Z = \frac{699 - 775}{75} = -1.20\)[/tex]
b. For $949:[tex]\(Z = \frac{949 - 775}{75} = 2.13\)[/tex]
c. For $625: [tex]\(Z = \frac{625 - 775}{75} = -2.33\)[/tex]
d. For $849: [tex]\(Z = \frac{849 - 775}{75} = 0.67\)[/tex]
e. For $999: [tex]\(Z = \frac{999 - 775}{75} = 3.13\)[/tex]
The z-score measures how many standard deviations a particular value is away from the mean.
A positive z-score indicates a value above the mean, while a negative z-score indicates a value below the mean.
The absolute value of the z-score tells you how many standard deviations the value is from the mean.
In this context, it helps you understand how each laptop price compares to the average price in terms of standard deviations.
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Prove each of the following statements below using one of the proof techniques and state the proof strategy you use.
a. The product of an odd and even integer is even.
b. Let m and n be integers. Show that if mn is even, then m is even or n is even.
c. If r is a nonzero rational number and p is an irrational number, then rp is irrational.
d. For all real numbers a, b,and c, max(a, max(b,c)) = max(max(a, b),c).
e. If a and bare rational numbers, then ab is rational too.
f. If a and bare two distinct rational numbers, then there exists an irrational number between them.
g. If m +n and n+p are even integers where m, n,p are integers, then m +p is even.
Answer:
See below
Step-by-step explanation:
a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.
b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.
c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.
d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:
a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b
Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.
e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.
f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.
g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.
A paper company needs to ship paper to a large printing business. The paper will be shipped in small boxes and large boxes. Each small box of paper weighs 45 pounds and each large box of paper weighs 80 pounds. There were twice as many large boxes shipped as small boxes shipped and the total weight of all boxes was 1435 pounds. Determine the number of small boxes shipped and the number of large boxes shipped.
Answer:
your... syllabus is different from mine....
Step-by-step explanation:
which class u study
I have four different sweaters. Every day I choose one of the four sweaters at random to wear. Let X be the number of different sweaters I wore during a 5-day week.
(For example, if my 5-day sweater sequence is 3,2,4,4,2 then X = 3 because that week I wore sweaters {2,3.4}.)
Find the mean of X.
The mean of the number of different sweaters worn in a week is 2.76. This is calculated by multiplying each possible outcome (from 1 to 4 different sweaters) by its probability and then adding these up.
Explanation:The random variable X here is the number of different sweaters you wear in a week. To find the mean (expected value) of a random variable, you need to take each possible outcome, multiply it by its probability, and then add up all these products.
Here are the possibilities:
You could wear the same sweater every day - there's 4 ways this can happen, so the probability is 4/(4^5) = 4/1024. You could wear 2 different sweaters - there are 4*3*2^4=192 ways this can happen, so the probability is 192/1024. You could wear 3 different sweaters - there are 4*3*2*3^2=432 ways this can happen, so the probability is 432/1024. You could wear 4 different sweaters - there are 4*3*2*1*4=96 ways this can happen, so the probability is 96/1024. You could wear all 4 different sweaters - there are 4! = 24 ways this can happen, so the probability is 24/1024.
So, the expected value or mean of X is:
1*(4/1024) + 2*(192/1024) + 3*(432/1024) + 4*(96/1024) + 4*(24/1024) = 2.76
So on an average week, you would expect to wear about 2 to 3 different sweaters.
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Following are the published weights (in pounds) of all of the team members of the Arizona Cardinals from a previous year. 177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265 Organize the data from smallest to largest value. When Jake Plummer, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he? Answer in the format .99 If your answer has a negative sign, enter it before the decimal.
Answer:
[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]
So then the value of 205 it's 0.76 deviations below the population mean on this case
Step-by-step explanation:
For this case we have the following data given:
177, 205, 210, 210, 232, 205, 185, 185, 178, 210, 206, 212, 184, 174, 185, 242, 188, 212, 215, 247, 241, 223, 220, 260, 245, 259, 278, 270, 280, 295, 275, 285, 290, 272, 273, 280, 285, 286, 200, 215, 185, 230, 250, 241, 190, 260, 250, 302, 265, 290, 276, 228, 265
And these values represent the weigths of all the team members of the Arizona Cardinals
Now if we organize the data values from the smallest to the largest we have:
174 177 178 184 185 185 185 185 188 190 200 205 205 206 210 210 210 212 212 215 215 220 223 228 230 232 241 241 242 245 247 250 250 259 260 260 265 265 270 272 273 275 276 278 280 280 285 285 286 290 290 295 302
For this case we can calculate the mean with the following formula:
[tex] \mu = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And if we replace we got:
[tex] \mu = 233.3396[/tex]
And for the deviation we can use the following formula:
[tex] \sigma =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And if we replace we got:
[tex] \sigma = 37.498[/tex]
And in order to calculate How many standard deviations above or below the mean was he we can use the z score formula given by:
[tex] z = \frac{x -\mu}{\sigma}[/tex]
And we assume that x=205 and if we replace we have:
[tex] z = \frac{205-233.3396}{37.498}= -0.76[/tex]
So then the value of 205 it's 0.76 deviations below the population mean on this case
Sven starts walking due south at 6 feet per second from a point 140 feet north of an intersection. At the same time Rudyard starts walking due east at 4 feet per second from a point 170 feet west of the intersection.
(a) Write an expression for the distance d between Sven and Rudyard t seconds after they start walking.
(b) When are Sven and Rudyard closest? (Round your answer to two decimal places.)
What is the minimum distance between them? (Round your answer to two decimal places.)
Answer:
a) y= y₀+vy*t , x= x₀+vx*t
b) they are closest at t= 29.23 s
c) r min = 63.79 ft
Step-by-step explanation:
a) denoting v as velocities and "₀" as initial conditions , then the position of Sven is given by the coordinate (0,y) where
y= y₀+vy*t
and the position of Rudyard is given by the coordinate (x,0) where
x= x₀+vx*t
b) the distance r between Sven and Rudyard is given by
r²=x²+y²
the distance will be minimum when the derivative of r with respect to the time is 0 . Then taking the derivative of the equation above
2*r*dr/dt = 2*x*dx/dt + 2*y*dy/dt
since dx/dt= vx and dy/dt= vy , then
r*dr/dt = x*vx+ y*vy
dr/dt = (x*vx+ y*vy)/r
assuming that r cannot be 0 , then
dr/dt =0 → x*vx+ y*vy = 0
(x₀+vx*t)*vx + (y₀+vy*t)*vy = 0
-(x₀*vx + y₀*vy) = (vx²+vy²)*t
t= -(x₀*vx + y₀*vy)/(vx²+vy²)
replacing values
t= -(x₀*vx + y₀*vy)/(vx²+vy²) = -[ 140 ft*(-6ft/s) + (-170 ft)*4 ft/s]/[ (-6ft/s)²+ (4 ft/s)²] = 29.23 s
then they are closest at t= 29.23 s
and the minimum distance will be
x = x₀ + vx*t = 140 ft+(-6ft/s)*29.23 s = -35.38 ft
y= y₀+vy*t = (-170 ft)+ 4 ft/s*29.23 s = -53.08 ft
r min = √(x²+y²)= 63.79 ft
r min = 63.79 ft
Note
to prove our assumption that r is not 0 , then x and y should be 0 at the same time. thus
0= y₀+vy*t → t = (-y₀)/vy = -140 ft/(-6ft/s) = 26.33 s
0= x₀+vx*t → t= (-x₀)/vx = -(-170 ft)/4 ft/s = 42.5 s
then r is never 0
Identify the type of data (qualitative/quantitative) and the level of measurement for the data described below. Explain your choice. The average monthly rainfall in inches for a certain city throughout the year Are the data qualitative or quantitative?
A. Quantitative, because numerical values, found by either measuring or counting, are used to describe the data
B. Qualitative, because numerical values, found by either measuring or counting, are used to describe the data °
C. Qualitative, because descriptive terms are used to measure or classify the data 0
D. Quantitative, because descriptive terms are used to measure or classify the data
What is the data set's level of measurement?
A. Nominal, because the data are categories or labels that cannot be ranked.
B. Ordinal, because the data are categories or labels that can be ranked °
C. Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.
D. Interval, because the differences in the data can be meaningfully measured, but the data do not have a true zero point.
Answer:
Option A) Quantitative, because numerical values, found by either measuring or counting, are used to describe the data
Option C) Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.
Step-by-step explanation:
We are given the following in the question:
"The average monthly rainfall in inches for a certain city throughout the year"
Qualitative or quantitative:
Quantitative data are measures of values or counts and are expressed as numbers. Qualitative data are measures of 'types' and may be represented by a name, symbol, or a number code. They are non-parametric values.Thus, the given quantitative data is quantitative because rainfall is always measured and the output is expressed in numerical values.
Option A) Quantitative, because numerical values, found by either measuring or counting, are used to describe the data
Level of measurement:
Ordinal: These are the qualitative variable whose order plays an important role.Interval: When true zero does not exist. The negative values of such variables make sense.Ratio: When true zero exist. The negative values of such variable does not make any sense.Since,for the given data true zero exist and a negative values make no sense, thus its is ratio.
Option C) Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.
The data described is quantitative because it is measured in numerical values. The level of measurement for this data is interval because the differences between data points can be meaningfully measured, but the data lacks a true zero point.
The data about the average monthly rainfall in inches for a certain city throughout the year is a Quantitative Data. This is because the data are numerical values which are obtained through measuring. Quantitative data is based on numbers and can be manipulated using statistical methods. The type of measurement used for this data set is Interval. This is because although the differences in the data (rainfall from one month to the next) can be meaningfully measured, the data do not have a 'true zero' point since there isn't a point where there’s absolutely no rainfall.
Learn more about Quantitative Data and Level of Measurement here:
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In a library where they atio of men to women is 1 to 1,
of the total members are men.
2
1/1
1/2
3
Answer:
1/2
Step-by-step explanation:
Well if we think logically it is quite simple.
Since there is a 1-in-1 ratio of men and women, this means that there are the same number of men and women.
Now taking a random value such as 6 employees, this means that you will see 3 men and 3 women, now in relation to the total it is 3/6
3/6 = 1/2