Answer:
Step-by-step explanation:
Hello!
The objective is to test if the mean burning rate of a propellant follows the specificated 50 cm/s.
The variable of interest is X: burning rate of a propellant (cm/s)
A sample of n= 25 was taken and a sample mean X[bar]= 51.3 cm/s and a sample standard deviation S= 2.0 cm/s were obtained.
Assuming that the variable has a normal distribution, the parameter of interest is the population mean, and the statistic hypotheses are:
H₀: μ = 50
H₁: μ ≠ 50
α: 0.05
Since there is no information about the population variance and the variable has a normal distribution the statistic to choose is a one-sample t-test:
[tex]t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{51.3-50}{\frac{2.0}{\sqrt{25} } } = 3.25[/tex]
The p-value corresponding to this test is p-value 0.003402
The decision for deciding using the p-value is:
If p-value ≤ α ⇒ Reject the null hypothesis.
If p-value > α ⇒ Not reject the null hypothesis.
The p-value is less than the significance level, the decision is to reject the null hypothesis. Using a level of significance of 5% you can conclude that the population mean burning rate of propellant is not 50 cm/s.
I hope it helps!
Consider randomly selecting a single individual and having that person test drive 3 different vehicles.
Define events A1, A2, and A3 by A1 = likes vehicle #1 A2 = likes vehicle #2 A3 = likes vehicle #3. Suppose that P(A1) = 0.55, P(A2) = 0.65, P(A3) = 0.70, P(A1 ∪ A2) = 0.80, P(A2 ∩ A3) = 0.50, and P(A1 ∪ A2 ∪ A3) = 0.88.
(a) What is the probability that the individual likes both vehicle #1
and vehicle #2?
(b) Determine and interpret P (A2 |A3 ).
(c) Are A2 and A3 independent events? Answer in two different ways.
(d) If you learn that the individual did not like vehicle #1, what now
is the probability that he/she liked at least one of the other two
vehicles?
Answer:
(a) The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40.
(b) The value of P (A₂ | A₃) is 0.7143.
(c) The events A₂ and A₃ are not independent.
(d) The probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.
Step-by-step explanation:
The events are defined as follows:
A₁ = an individual like vehicle #1
A₂ = an individual like vehicle #2
A₃ = an individual like vehicle #3
The information provided is:
[tex]P(A_{1})=0.55\\P(A_{2})=0.65\\P(A_{3})=0.70\\P(A_{1}\cup A_{2})=0.80\\P(A_{2}\cap A_{3})=0.50\\P(A_{1}\cup A_{2}\cip A_{3})=0.88\\[/tex]
(a)
Compute the probability that the individual likes both vehicle #1 and vehicle #2 as follows:
[tex]P(A_{1}\cap A_{2})=P(A_{1})+P(A_{2})-P(A_{1}\cup A_{2})\\=0.55+0.65-0.80\\=0.40[/tex]
Thus, the probability that the individual likes both vehicle #1 and vehicle #2 is 0.40.
(b)
Compute the value of P (A₂ | A₃) as follows:
[tex]P(A_{2}|A_{3})=\frac{P(A_{2}\cap A_{3})}{P(A_{3}}\\=\frac{0.50}{0.70}\\=0.7143[/tex]
Thus, the value of P (A₂ | A₃) is 0.7143.
(c)
If two events X and Y are independent then,
[tex]P(X\cap Y)=P(X)\times P(Y)\\P(X|Y)=P(X)[/tex]
The value of P (A₂ ∩ A₃) is 0.50.
The product of the probabilities, P (A₂) and P (A₃) is:
[tex]P(A_{2})\times P(A_{3})=0.65\times0.70=0.455[/tex]
Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃)
The value of P (A₂ | A₃) is 0.7143.
The value of P (A₂) is 0.65.
Thus, P (A₂ | A₃) ≠ P (A₂).
The events A₂ and A₃ are not independent.
(d)
Compute that probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ as follows:
[tex]P(A_{2}\cup A_{3}|A_{1}^{c})=\frac{P((A_{2}\cup A_{3})\cap A_{1}^{c})}{P(A_{1}^{c})}\\=\frac{P((A_{2}\cup A_{3}\cup A_{1})-P(A_{1})}{1-P(A_{1})} \\=\frac{0.88-0.55}{1-0.55}\\=0.7333[/tex]
Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.
Final answer:
The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40. Events A2 and A3 are not independent as P(A2 and A3) does not equal P(A2)P(A3). If the individual did not like vehicle #1, the probability of liking at least one of the other two vehicles is 0.73.
Explanation:
Probability of Likes and Independency of Events
To find the probability that the individual likes both vehicle #1 and vehicle #2, which is P(A1 ∩ A2), we use the formula: P(A1 ∩ A2) = P(A1) + P(A2) - P(A1 ∪ A2). Given that P(A1) = 0.55, P(A2) = 0.65, and P(A1 ∪ A2) = 0.80, we can calculate P(A1 ∩ A2):
P(A1 ∩ A2) = 0.55 + 0.65 - 0.80 = 0.40.
The conditional probability P(A2|A3) represents the probability of liking vehicle #2 given that the individual already likes vehicle #3. Using the definition of conditional probability:
P(A2|A3) = P(A2 ∩ A3) / P(A3) = 0.50 / 0.70 = approximately 0.714.
To determine if A2 and A3 are independent events, we can check if P(A2 ∩ A3) equals P(A2)P(A3) which would be 0.65 * 0.70 = 0.455. Since P(A2 ∩ A3) = 0.50, which is not equal to 0.455, A2 and A3 are not independent.
Moreover, if A2 and A3 were independent, then knowing that A3 occurred would not change the probability of A2 happening. However, P(A2|A3) is not equal to P(A2), indicating their dependence. Lastly, when the individual did not like vehicle #1, the probability that they liked at least one of the other two vehicles can be found using the probability of the union of events, subtracting the probability of liking vehicle #1:
P(A2 ∪ A3 | not A1) = P(A2 ∪ A3) - P(A1) + P(A1 ∩ A2 ∩ A3) = 0.88 - 0.55 + 0.40 = 0.73.
What is 3 1/4 +( 3 1/4 +5 1/5)?
Answer:16.5
Step-by-step explanation:
Answer:
11 7/10
Step-by-step explanation:
Lets make the fractions into common denominators, by mulitplying 4 by 5 and 5 by 4 (see below) and follow PEMDAS
3 1/4(5) +(3 1/4(5) + 5 1/5(4))
3 5/20+ (3 5/20+5 4/20)
Simplify like terms
3 5/20+ (8 9/20)
11 14/20
Simplify fraction
11 7/10
. You decide you want to start investing for your retirement and you want to have $650,000. If you save $100 a month in an account that averages a 10% annual rate of return, will you have enough money in 25 years? (Hint: this formula was introduced in Section 1)
Answer:
750,0000
Step-by-step explanation:
how many pounds of chamomile tea that costs 18.20 per pound must be mixed with 12lb of orange tea that costs 12.25 per pound to make a mixture that costs 14.63 per pound
8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
Step-by-step explanation:
We are given the following in the question:
Chamomile tea:
Unit cost = 18.20 per pound
Amount = x pounds
Total cost =
[tex]18.20\times x = 18.20x[/tex]
Orange tea:
Unit cost = 12.25 per pound
Amount = 12 pounds
Total cost =
[tex]12.25\times 12 = 147[/tex]
Total mixture:
Unit cost = 14.63 per pound
Amount = (12+x) pounds
Total cost =
[tex]14.63\times (12+x) = 175.56 + 14.63x[/tex]
We can write the equation that cost of mixture is equal to cost of chamomile tea and orange tea.
[tex]18.20x + 147 = 175.56 + 14.63x\\\Rightarrow 18.20x- 14.63x = 175.56-147\\\Rightarrow 3.57x = 28.56\\\Rightarrow x = 8[/tex]
Thus, 8 pounds of chamomile tea must be mixed to make a mixture that costs 14.63 per pound.
Imagine that you have been given a dataset of 1,000 documents that have been classified as being about entertainment or education. There are 700 entertainment documents in the dataset and 300 education documents in the dataset. The tables below give the number of documents from each topic that a selection of words occurred in. Word-document counts for the entertainment dataset fun christmas is 695 machine 35 family 400 learning 70 415 Word-document counts for the education dataset is christmas fun 200 machine 120 family 10 learning 105 295a. What target level will a naive Bayes model predict for the following query document: "machine learning is fun"? b. What target level will a naive Bayes model predict for the following query document: "christmas family fun"? c. What target level will a naive Bayes model predict for the query document in part (b) of this question, if Laplace smoothing with k = 10 and a vocabulary size of 6 is used?
Answer:
Please find attached answer.
Step-by-step explanation:
The Naive Bayes model is likely to classify the document 'machine learning is fun' and 'christmas family fun' as 'entertainment', even when using Laplace smoothing.
Explanation:The Naive Bayes model classifies texts based upon the likelihood of a particular term belonging to a given class. To find the target level for the document, you multiply the prior probability of the document being in a given class (either entertainment or education) with the likelihoods of the terms in the query document being in that class and you choose the class with the highest probability.
For the query document "machine learning is fun": Given the word-document counts, the model would predict 'entertainment' due to the higher frequency of 'fun' and 'machine' in entertainment documents than in education documents.For the query document "christmas family fun": The model would probably predict 'entertainment' due to the higher frequency of 'fun' and 'family' in the entertainment dataset than in the education dataset.If Laplace smoothing with k = 10 and a vocabulary size of 6 is used, the model would still predict 'entertainment' for "christmas family fun", as adjusting for unseen words in this way will not change the relative probabilities significantly.Learn more about Naive Bayes model here:https://brainly.com/question/21507963
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The Christmas Bird Count (CBC) is an annual tradition in Lexington, Massachusetts. A group of volunteers counts the number of birds of different species over a 1-day period. Each year, there are approximately 30–35 hours of observation time split among multiple volunteers. The following counts were obtained for the Northern Cardinal (or cardinal, in brief) for the period 2005–2011.
Year Number
2005 76
2006 47
2007 63
2008 53
2009 62
2010 69
2011 62
5.126 What is the mean number of cardinal birds per year observed from 2005 to 2011?
5.127 What is the standard deviation (sd) of the number of cardinal birds observed?
5.128 What is the probability of observing at least 60 cardinal birds in 2012? (Hint: Apply a continuity correction where appropriate.)
The observers wish to identify a normal range for the number of cardinal birds observed per year. The normal range will be defined as the interval (L, U), where L is the largest integer ≤ 15th percentile and U is the smallest integer ≥ 85th percentile.
5.129 If we make the same assumptions as in Problem 5.128, then what is L? What is U?
5.130 What is the probability that the number of cardinal birds will be ≥ U at least once on Christmas day during the 10-year period 2012–2021? (Hint: Make the same assumptions as in Problem 5.128.)
Answer:
(5.216) mean = 61.71
(5.217) standard deviation = 8.88
(5.218) P(X>=60) = 0.9238
(5.129) L = 79, U = 69
(5.130) P(X> or = U) = 0.7058
Step-by-step explanation:
The table of the statistic is set up as shown in attachment.
(5.216) mean = summation of all X ÷ no of data.
mean = 432/7 = 61.71 birds
(5.217)Standard deviation = √ sum of the absolute value of difference of X from mean ÷ number of data
S = √ /X - mean/ ÷ 7
= √551.428/7
S = 8.88
(5.218) P (X> or = 60)
= P(Z> or =60 - 61.71/8.8 )
= P(Z>or= - 0.192)
= 1 - P(Z< or = 0.192)
= 1- 0.0762
= 0.9238
(5.219)the 15th percentile=15/100 × 7
15th percentile = 1.05
The value is the number in the first position and that is 79,
L= 79
85th percentile = 85/100 × 7 = 5.95
The value is the number in the 6th position, and that is 69
U = 69
5.130) P(X>or = 60)
= P(Z>or= 69 - 61.71/8.8)
= P(Z> or = 0.8208)
= 1 - P(Z< or = 0.8209)
= 1 - 0.2942
= 0.7058
Rona drove 56 milesto visit a friend she drove 42 miles before stopping for gas what percent of the drive did rona complete before stopping for gas
Answer:
75%
Step-by-step explanation:
Find the percentage corresponding to 42/56: (0.75)(100%) = 75%
Answer:
75%
Step-by-step explanation:
Becauuuuuuseeee if you divide 42 by 56 it would be 0.75 and yeah. so take out the 0 and its 75%
ALGEBRA 2 HELP!!!
Find all the linear factors
f(x)=[tex]3x^{4}-7x^{3}-10x^{2} +28x-8[/tex]
Answer:
x + 2
3x − 1
x − 2
Step-by-step explanation:
The order of the polynomial is 4, so there are 4 roots.
Use rational root theorem to find possible rational roots.
±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3
By trial and error, -2, ⅓, and 2 are all roots. The fourth root can't be imaginary (complex roots come in pairs), so one of the real roots must be repeated.
Answer: the factors are (x - 2)(x + 2)(3x - 1)
Step-by-step explanation:
The given polynomial function is expressed as
f(x) = 3x⁴ - 7x³ - 10x² + 28x - 8
The first step is to test for any value of x that satisfies the polynomial when
3x⁴ - 7x³ - 10x² + 28x - 8 = 0
Assuming x = 2, then
3(2)⁴ - 7(2)³ - 10(2)² + 28 × 2 - 8 = 0
48 - 56 - 40 + 56 - 8 = 0
0 = 0
It means that x - 2 is a factor.
To determine the other factors, we would apply the long division method. The steps are shown in the attached photo. Looking at the photo, we would factorize the quadratic equation which is expressed as
3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x + 2) - 1(x + 2)
(3x - 1)(x + 2) = 0
Suppose 10000 people are given a medical test for a disease. About1% of all people have this condition. The test results have a 15% false positive rate and a 10% false negative rate. What percent of the people who tested positive actually have the disease?
Answer:
The percent of the people who tested positive actually have the disease is 38.64%.
Step-by-step explanation:
Denote the events as follows:
X = a person has the disease
P = the test result is positive
N = the test result is negative
Given:
[tex]P(X)=0.01\\P(P|X^{c})=0.15\\P(N|X)=0.10[/tex]
Compute the value of P (P|X) as follows:
[tex]P(P|X)=1-P(P|X^{c})=1-0.15=0.85[/tex]
Compute the probability of a positive test result as follows:
[tex]P(P)=P(P|X)P(X)+P(P|X^{c})P(X^{c})\\=(0.85\times0.10)+(0.15\times0.90)\\=0.22[/tex]
Compute the probability of a person having the disease given that he/she was tested positive as follows:
[tex]P(X|P)=\frac{P(P|X)P(X)}{P(P)}=\frac{0.85\times0.10}{0.22} =0.3864[/tex]
The percentage of people having the disease given that he/she was tested positive is, 0.3864 × 100 = 38.64%.
Suppose that, during the next hour, 20 customers each purchase one cup of coffee. Assume that the customers make their decisions independently, and each customer has a 17% probability of ordering decaffeinated coffee. What is the probability that everyone will order a coffee with caffeine? g
Answer:
0.0241 or 2.41%
Step-by-step explanation:
Since each customer has a 17% probability of ordering decaffeinated coffee, the probability of a customer ordering a coffee with caffeine is:
[tex]P(C) = 1-0.17 = 0.83[/tex]
If customers make their decisions independently, the probability that all 20 order a coffee with caffeine is:
[tex]P(X=20)=P(C)^{20}\\P(X=20) = 0.83^{20}=0.0241=2.41\%[/tex]
The probability is 0.0241 or 2.41%.
The area of a rectangle is 18 square inches. The length of the rectangle is twice its width. Find the width of the rectangle (in inches).
Let the width = X
Then the length would be 2X ( twice the width).
Area is Length x width, so you have X * 2X = 18
x *2x = 2x^2
Now you have 2x^2 = 18
Divide both sides by 2:
x^2 = 9
Take the square root of both sides:
x = sqrt(9)
x = 3
The width is 3 inches.
The width of the rectangle is determined by establishing an equation from the given area (18 square inches) and the fact that the length is twice the width, which leads to the solution that the width is 3 inches.
Explanation:To find the width of a rectangle when the area is 18 square inches, and the length is twice the width, we need to set up an equation. Let the width be w inches. Then the length would be 2w inches, since it's twice the width. We know that the area (A) of a rectangle is the multiplication of its length and width, so we have:
A = length imes width
18 = 2w imes w
18 = 2w2
Now we solve for w:
w2 = 18 / 2
w2 = 9
w = 3
The width of the rectangle is 3 inches.
A publisher reports that 344% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 220220 found that 300% of the readers owned a particular make of car. Is there sufficient evidence at the 0.020.02 level to support the executive's claim
Answer:
No, there is not enough evidence at the 0.02 level to support the executive's claim.
Step-by-step explanation:
We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.
And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;
Null Hypothesis, [tex]H_0[/tex] : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}
Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}
The test statistics we will use here is;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, p = actual % of readers who own a particular make of car = 0.34
[tex]\hat p[/tex] = percentage of readers who own a particular make of car in a
sample of 220 = 0.30
n = sample size = 220
So, Test statistics = [tex]\frac{0.30 -0.34}{\sqrt{\frac{0.30(1- 0.30)}{220} } }[/tex]
= -1.30
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.
A state insurance commission estimates that 13% of all motorists in its state areuninsured. Suppose this proportion is valid. Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured. You may assume that the normaldistribution applies.
Answer:
79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
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.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 50, p = 0.13[/tex]
So
[tex]\mu = E(X) = np = 50*0.13 = 6.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.13*0.87} = 2.38[/tex]
Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
This is 1 subtracted by the pvalue of Z when X = 4. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 6.5}{2.38}[/tex]
[tex]Z = -0.84[/tex]
[tex]Z = -0.84[/tex] has a pvalue of 0.2005
1 - 0.2005 = 0.7995
79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.
Using the normal approximation to the binomial distribution, the probability that at least 5 out of 50 motorists will be uninsured when 13% of motorists are uninsured is approximately 74.22%.
First, we calculate the mean (μ) and the standard deviation (σ) of the number of uninsured motorists in a sample of 50. The mean number of uninsured motorists is μ = np, where n is the sample size and p is the probability of a motorist being uninsured. So, μ = 50 * 0.13 = 6.5.
The standard d eviation is σ = √(np(1-p)). Therefore, σ = √(50 * 0.13 * (1 - 0.13)) ≈ 2.31.
To find the probability of at least 5 uninsured motorists, we want P(X ≥ 5). We need to standardize this to the standard normal distribution to find the z-score. The z-score is z = (X - μ) / σ.
For X = 5, the z-score is z = (5 - 6.5) / 2.31 ≈ -0.65. We use the z-table or a calculator to find the probability corresponding to z = -0.65.
Since we want at least 5 uninsured, we find P(Z ≥ -0.65), which equals 1 - P(Z < -0.65). The value from the z-table for -0.65 is approximately 0.2578. Therefore, the probability we're looking for is 1 - 0.2578 = 0.7422, or 74.22%.
Consider the infinite series sigma ^infinity _k = 1 1/7^k a. Find the first four terms of the sequence of partial sums. b. Use the results of part (a) to propose a formula for S_n. c. Propose a value of the series. a. Find the first four terms of the sequence of partial sums. S_1 =, S_2 =, S_3 =, S_4 = (Simplify your answers.) b. Use the results of part (a) to propose a formula for S_n. A. S_n = 7^n - 1/6 middot 7^n B. S_n = 7n/7^n C. S_n = n/7^n D. S_n = 6 middot 7^n/7^n - 1 c. Propose a value of the series. A reasonable conjecture for the value of the series is
Answer:
a. First four terms of the sequential partial sums
[tex]S_1=\frac{1}{7}, S_2=\frac{7^2-1}{6*7^2}, S_3 =\frac{7^3-1}{6*7^3}, S_4 =\frac{7^4-1}{6*7^4}[/tex]
b. The formula for Sn is option A = [tex]\frac{7^n-1}{6*7^n}[/tex]
c. Value of the series = [tex]\frac{1}{6}[/tex]
Step-by-step explanation:
a. First four terms of the sequential partial sums
[tex]\sum^{\infty}_{k=1}\\S_1=\frac{1}{7}\\ S_2=\frac{1}{7}+ \frac{1}{7^2}= \frac{8}{49} =\frac{7^2-1}{6*7^2}\\S_3=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3} = \frac{57}{343} =\frac{7^3-1}{6*7^3}\\S_4=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4} = \frac{400}{2401} =\frac{7^4-1}{6*7^4}[/tex]
b. Formula for Sn
The sum of n terms
[tex]S_n=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+ \frac{1}{7^4}+.....+ \frac{1}{7^n}= \frac{7^n-1}{6*7^n}[/tex]
c. Value for the series
This can be given as the Sum of infinite terms;
[tex]S_{\infty}=\frac{1}{7}+ \frac{1}{7^2}+ \frac{1}{7^3}+ \frac{1}{7^4}+....+ \frac{1}{7^n}+ ....= \lim_{n \to \infty} \frac{7^n-1}{6*7^n}= \frac{1}{6}[/tex]
A recent survey of 1010 U.S. adults selected at random showed that 627 consider the occupation of firefighter to have very great prestige. Estimate the probability (to the nearest hundredth) that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.
Answer:
0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Desired outcomes:
Adults who consider the occupation of firefighter to have very great prestige. So the number of desired outcomes is [tex]D = 627[/tex]
Total outcomes.
All adults sampled. So [tex]T = 1010[/tex]
Probability:
[tex]P = \frac{D}{T} = \frac{627}{1010} = 0.62[/tex]
0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night. Assume that room rates are normally distributed with a standard deviation of $55. What is the minimum cost that captures the 20% most expensive hotel rooms in New York City?
Answer:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 250.31.
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 room hotel rate of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(204,55)[/tex]
Where [tex]\mu=204[/tex] and [tex]\sigma=55[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.20[/tex] (a)
[tex]P(X<a)=0.80[/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 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.842[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.842[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.842<\frac{a-204}{55}[/tex]
And if we solve for a we got
[tex]a=204 +0.842*55=250.31[/tex]
So the value of height that separates the bottom 80% of data from the top 20% is 250.31.
Jon is hitting baseballs. When he tosses the ball into the air, his hand is 5 feet above the ground. He hits the ball when it falls back to a height of 4 feet. The height of the ball is given by h = 5 + 25t - 16t^2, where h is measured in feet and t is measured in seconds. How many seconds after throwing the ball does Jon hit it?
Answer:
[tex]t=1.60\ sec[/tex]
Step-by-step explanation:
Functions
We are given the function that relates the height of the ball and the time elapsed t. The function is:
[tex]h = 5 + 25t - 16t^2[/tex]
We need to know the time needed for the height to reach 4 feet, that is h=4
[tex]5 + 25t - 16t^2=4[/tex]
Rearranging
[tex]16t^2-25t -1=0[/tex]
Solving for t
[tex]\boxed{t=1.60\ sec}[/tex]
The other solution is negative and is discarded because t cannot be negative.
A fair die is rolled 8 times. What is the probability that the die comes up 6 exactly twice? What is the probability that the die comes up an odd number exactly five times? Find the mean number of times a 6 comes up. Find the mean number of times an odd number comes up. Find the standard deviation of the number of times a 6 comes up. Find the standard deviation of the number of times an odd number comes up.
Answer:
0.2605, 0.2188, 1.33, 4, 1.0540, 1.4142
Step-by-step explanation:
A fair die is rolled 8 times.
a. What is the probability that the die comes up 6 exactly twice?
b. What is the probability that the die comes up an odd number exactly five times?
c. Find the mean number of times a 6 comes up.
d. Find the mean number of times an odd number comes up.
e. Find the standard deviation of the number of times a 6 comes up.
f. Find the standard deviation of the number of times an odd number comes up.
a. A die is rolled 8 times. If A represent the number of times a 6 comes up. For a fair die the probability that the die comes up 6 is 1/6 - Thus A ~ Bin(8, 1/6)
The probability mass function of the random variable A is
[tex]p(A) = \left \{ {\frac{8!}{x!(8 - x)!}*(\frac{1}{6} )^{A}*(\frac{5}{6} )^{8-A} } \right. for A=0,1, ...8[/tex]
hence, p(6 twice) implies P(A=2)
that is P(2) substitute A = 2
[tex]p(2) = \left \{ {\frac{8!}{2!(8 - 2)!}*(\frac{1}{6} )^{2}*(\frac{5}{6} )^{8-2} } \right. for A=0,1, ...8[/tex]
[tex]p(2)=\frac{8!}{2!6!} *(\frac{1}{6} )^{2} *(\frac{5}{6} )^{6}[/tex]
p(2) = 0.2605
b. If B represent the number of times an odd number comes up. For the fair die the probability that an odd number comes up is 0.5.
Thus B ~ Bin(8, 1/2 )
The probability mass function of the random variable B is given by
[tex]p(B) = \left \{ {\frac{8!}{B!(8 - B)!}*(\frac{1}{2} )^{B\\}*(\frac{1}{2} )^{8-B} } \right. for B=0,1, ...8[/tex]
hence p(odd comes up 5 times) is
[tex]p(x=5) = p(2)=\frac{8!}{5!3!} *(\frac{1}{2} )^{5} *(\frac{1}{2} )^{3}[/tex]
p(5) = 0.2188
c. let the mean no of times a 6 comes up be μₐ
and let the total number of outcomes be n
using the formula μₐ = nρₐ
μₐ = 8 * 1/6
= 1.33
d. let the mean nos of times an odd nos comes up be μₓ
let the total outcomes be n = 8
let the probability odd be pb = 1/2
μₓ = npb
= 8 * (1/2)
= 4
e. the standard deviation of a random variable A is given as follows
σₐ [tex]= \sqrt{np(1-p)}[/tex]
where p = 1/6 (prob 6 outcome)
n = total outcomes = 8
[tex]= \sqrt{8*\frac{1}{6}*\frac{5}{6} }[/tex]
= 1.0540
f. the standard dev of the binomial random variable Y is given by
σ [tex]= \sqrt{np(1-p)}[/tex]
where p = 1/2 and n = 8
= [tex]\sqrt{8*\frac{1}{2} *\frac{1}{2} }[/tex]
= 1.4142
In a study of cereal leaf beetle damage to oats, researchers measured the number of beetle larvae per stem in small plots of oats after applying (or not applying) the pesticide Malathion . Researchers applied Malathion to a random sample of 5 plots, and did not apply the pesticide to an independent sample of 12 plots. A noted scientist claims that Malathion will not make any difference in the mean number of larvae per stem. Test her claim at the .05 level of significance. State H0 and H1. a. H0: Mean Malathion -MeanNoMalathion <= 0 H1: Mean Malathion-Mean Malathion > 0 b. H0: MeanMalathion -MeanNoMalathion >= 0 H1: MeanMalathion-MeanNoMalathion < 0 c. H0: MeanMalathion = MeanNoMalathion H1: MeanMalathion ≠ MeanNoMalathion d. H0: MeanMalathion-MeanNoMalathion = 0 H1: MeanMalathion-MeanNoMalathion ≠ 0
Answer:
At 0.05 level of significance, there is no difference in the mean number of larvae per stem. This supports the scientist's claim.
(d) H0: MeanMalathion - MeanNoMalathion equals 0
H1: MeanMalathion - MeanNoMalathion not equals 0
Step-by-step explanation:
Test statistic (t) = (mean 1 - mean 2) ÷ sqrt[pooled variance (1/n1 + 1/n2)]
Let the difference between the two means be x and the pooled variance be y
n1 = 5, n2 = 12
t = x ÷ sqrt[y(1/5 + 1/12)] = x ÷ sqrt(0.283y) = x ÷ 0.532√y = 1.88x/√y
Assuming the ratio of x to √y is 0.5
t = 1.88×0.5 = 0.94
n1 + n2 = 5 + 12 = 17
degree of freedom = n1 + n2 - 2 = 17 - 2 = 15
significance level = 0.05 = 5%
critical value corresponding to 15 degrees of freedom and 5% confidence interval is 2.131
The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.
The region of no rejection of the null hypothesis lies between -2.131 and 2.131
Conclusion
Fail to reject the null hypothesis because the test statistic 0.94 falls within the region bounded by the critical values.
The scientist's claim is right.
A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.
An alternate hypothesis is also a statement from a population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.
An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.
a) Write out the hypotheses.
b) Calculate the test statistic.
c) Find the p-value.
d) Do we reject the null hypothesis? Explain.
e) What can you conclude based on this evidence
Answer:
An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.
a) Write out the hypotheses.
Fewer than 75% of claims settled within two months of filing.
b) Calculate the test statistic.
Test statistic = percent of claims settled in two months = 23/35 = 65.7%
c) Find the p-value.
we need to use the z-score with a table
=
standard deviation = s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}, =
√(1/34) [(0.657)(12) + (0.343)(23)] = √0.463911765 = 0.681
let 1 = "solved" and 0 = "unresolved"
thus our mean is 0.657
z = (0.75 - 0.657)/(0.681) - 0.137
p = 0.55172
d) Do we reject the null hypothesis? Explain.
Yes, because our p value is not below 0.05 and is not substantial to prove our null hypothesis
e) What can you conclude based on this evidence?
That further testing is needed.
Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1511 grams and standard deviation 198 grams. Use the TI-84 Plus calculator to answer the following. (a) What proportion of broilers weigh between 1143 and 1242 grams?
Answer:
5.55% of broilers weigh between 1143 and 1242 grams
Step-by-step explanation:
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.
In this problem, we have that:
[tex]\mu = 1511, \sigma = 198[/tex]
What proportion of broilers weigh between 1143 and 1242 grams?
This is the pvalue of Z when X = 1242 subtracted by the pvalue of Z when X = 1143.
X = 1242
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1242 - 1511}{198}[/tex]
[tex]Z = -1.36[/tex]
[tex]Z = -1.36[/tex] has a pvalue of 0.0869
X = 1143
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1143 - 1511}{198}[/tex]
[tex]Z = -1.36[/tex]
[tex]Z = -1.86[/tex] has a pvalue of 0.0314
0.0869 - 0.0314 = 0.0555
5.55% of broilers weigh between 1143 and 1242 grams
3.3.1. An urn contains five balls numbered 1 to 5. Two balls are drawn simultaneously. (a) Let X be the larger of the two numbers drawn. Find pX (k). (b) Let V be the sum of the two numbers drawn. Find pV (k).
Answer:
Step-by-step explanation:
Given that an urn contains five balls numbered 1 to 5
Two balls are drawn simultaneously.
a) X = the larger of two numbers drawn
Assuming balls are drawn without replacement, (since simultaneously drawn)
the sample space would be (1,2) (1,3)(1,4)(1,5) (2,3)(2,4)(2,5) (3,4) (3,5) (4,5)
n(S) = 10
n(x=2) = 1,
n(x=3) =2
n(x=4) = 3
n(x=5) = 4
Larger value X can take values only as 2,3,4 or 5
X 2 3 4 5
p 0.1 0.2 0.3 0.4
-------------------------------------------
b) V = sum of numbers drawn
V can take values as 3, 4, 5, 6, 7, 8 or 9
V 3 4 5 6 7 8 9
p 0.1 0.1 0.2 0.2 0.2 0.1 0.1
The probability that the larger number drawn will be k follows a specific probability distribution, as does the sum of the two numbers drawn. Using the given information and principles of probability, we can determine the probabilities for each value of k for both X and V. For X, the probability distribution pX(k) is [1/5, 1/5, 3/10, 1/5, 1/5]. For V, the probability distribution pV(k) is [1/20, 1/10, 3/20, 1/10, 1/10, 1/10, 3/20, 1/10, 1/20].
(a) Let's consider each possible outcome to find pX(k), the probability that the larger number drawn will be k:
If k is 1, there is only one possible outcome: drawing (1, anything). The probability of this outcome is (1/5) * (4/4) = 1/5.
If k is 2, there are two possible outcomes: drawing (2, 1) or (2, 3). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/10 + 1/10 = 1/5.
If k is 3, there are three possible outcomes: drawing (3, 1), (3, 2), or (3, 4). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (2/4) = 1/10 + 1/10 + 1/10 = 3/10.
If k is 4, there are four possible outcomes: drawing (4, 1), (4, 2), (4, 3), or (4, 5). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (3/4) = 1/10 + 1/10 + 1/10 + 3/20 = 1/5.
If k is 5, there are five possible outcomes: drawing (5, 1), (5, 2), (5, 3), (5, 4), or (5, 5). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (4/4) = 1/10 + 1/10 + 1/10 + 1/10 + 1/5 = 1/5.
Therefore, the probability distribution pX(k) is:
pX(1) = 1/5
pX(2) = 1/5
pX(3) = 3/10
pX(4) = 1/5
pX(5) = 1/5
(b) Let's consider each possible outcome to find pV(k), the probability that the sum of the two numbers drawn will be k:
If k is 2, there is only one possible outcome: drawing (1, 1). The probability of this outcome is (1/5) * (1/4) = 1/20.
If k is 3, there are two possible outcomes: drawing (1, 2) or (2, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 = 1/10.
If k is 4, there are three possible outcomes: drawing (1, 3), (2, 2), or (3, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 = 3/20.
If k is 5, there are four possible outcomes: drawing (1, 4), (2, 3), (3, 2), or (4, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 = 1/10.
If k is 6, there are five possible outcomes: drawing (1, 5), (2, 4), (3, 3), (4, 2), or (5, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 + 1/20 = 1/10.
If k is 7, there are four possible outcomes: drawing (2, 5), (3, 4), (4, 3), or (5, 2). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 = 1/10.
If k is 8, there are three possible outcomes: drawing (3, 5), (4, 4), or (5, 3). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 = 3/20.
If k is 9, there are two possible outcomes: drawing (4, 5) or (5, 4). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 = 1/10.
If k is 10, there is only one possible outcome: drawing (5, 5). The probability of this outcome is (1/5) * (1/4) = 1/20.
Therefore, the probability distribution pV(k) is:
pV(2) = 1/20
pV(3) = 1/10
pV(4) = 3/20
pV(5) = 1/10
pV(6) = 1/10
pV(7) = 1/10
pV(8) = 3/20
pV(9) = 1/10
pV(10) = 1/20
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One of four calculator batteries is bad. An experiment consists of testing each battery until the dead one is found.
(a) How many possible outcomes are there for this experiment?
(b) Is the outcome GBGG (Good, Bad, Good, Good) possible? Why or why not?
Answer:
(a) 4 possible outcomes
(b) Not possible, testing stops upon finding the bad battery,
Step-by-step explanation:
Let G denote a good battery and B denote a bad battery.
Whenever a bad battery is found, the batteries stop being tested.
(a) The outcomes for this experiment are:
B
GB
GGB
GGGB
There are 4 possible outcomes.
(b) The outcome GBGG is not possible since testing stops upon finding the bad battery, the outcome in this case would be GB.
You are testing the null hypothesis that the population proportion equals .45, using data you collected from a sample of 100 adults. You sample proportion equals .30. What does Z equal
Answer:
[tex]z=\frac{0.3 -0.45}{\sqrt{\frac{0.45(1-0.45)}{100}}}=-3.015[/tex]
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
[tex]\hat p=0.3[/tex] estimated proportion of interest
[tex]p_o=0.45[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is 0.45.:
Null hypothesis:[tex]p=0.45[/tex]
Alternative hypothesis:[tex]p \neq 0.45[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.3 -0.45}{\sqrt{\frac{0.45(1-0.45)}{100}}}=-3.015[/tex]
The z-value is approximately -3.01.
To test the null hypothesis about a population proportion, you can use the z-test for proportions. The formula for the z-test statistic for proportions is:
[tex]z= \frac{p -p_0}{\frac{\sqrt{p_0(1 -p_0)}}{n} }[/tex]
where:
p is the sample proportion,
p_0 is the hypothesized population proportion under the null hypothesis, n is the sample size.
In your case:
p =0.30 (sample proportion),
p_0 =0.45 (hypothesized population proportion),
n=100 (sample size).
Now plug these values into the formula:
Calculate the values within the brackets first:
[tex]z= \frac{0.30 - 0.45}{\frac{\sqrt{0.45(1 -0.45)}}{100} }[/tex]
[tex]z= \frac{-0.15}{\frac{\sqrt{0.45(0.55)}}{100} }[/tex]
[tex]z= \frac{-0.15}{\frac{\sqrt{0.2475}}{100} }[/tex]
[tex]z= \frac{-0.15}{\sqrt{0.002475} }[/tex]
z≈−3.01
So, the z-value is approximately -3.01.
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Is (2,7) a point on the line y=4x-3?
Given that the equation of the line is [tex]y=4x-3[/tex]
We need to determine the point (2,7) lies on the line.
For a point to lie on the line, substituting the point in the equation of the line makes it valid.
Thus, substituting the point (2,7) in the equation of the line [tex]y=4x-3[/tex], we get;
[tex]7=4(2)-3[/tex]
[tex]7=8-3[/tex]
[tex]7\neq 5[/tex]
Thus, the both sides of the equation are not equal.
Substituting the point (2,7) in the equation makes the equation invalid.
Thus, the point (2,7) does not lie on the line [tex]y=4x-3[/tex]
Give (in percents) the two marginal distributions, for marital status and for income. Do each of your two sets of percents add to exactly 100%? If not, why not?
Answer:
Answer is attached
Step-by-step explanation:
6.11
Marginal Distribution of Marital Status
Single 4.1%
Married 93.9%
Divorced 1.5%
Widowed .5%
Marginal Distribution of Job Grade
Job Grade 1 11.6%
Job Grade 2 51.5%
Job Grade 3 30.2%
Job Grade 4 6.7%
Each of the marginal distributions sum up to exactly 100%. Depending on how one rounds the percentages, it is possible that the results will not be exactly 100%.
Test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.
The null and alternative hypothesis would be:
a. H0:p≥0.9H0:p≥0.9
H1:p<0.9H1:p<0.9
b. H0:p=0.9H0:p=0.9
H1:p≠0.9H1:p≠0.9
c. H0:p≤0.9H0:p≤0.9
H1:p>0.9H1:p>0.9
d. H0:μ=0.9H0:μ=0.9
H1:μ≠0.9H1:μ≠0.9
e. H0:μ≤0.9H0:μ≤0.9
H1:μ>0.9H1:μ>0.9
f. H0:μ≥0.9H0:μ≥0.9
H1:μ<0.9H1:μ<0.9
The test is:
a. left-tailed
b. two-tailed
c. right-tailed
Based on a sample of 100 people, 94% owned cats
The p-value is: (to 2 decimals)
Based on this we:
Fail to reject the null hypothesis
Reject the null hypothesis
Box 1: Select the best answer
Box 2: Select the best answer
Box 3: Enter your answer as an integer or decimal number. Examples: 3, -4, 5.5172
Enter DNE for Does Not Exist, oo for Infinity
Box 4: Select the best answer
Answer:
1. H0 : p = 0.9
H1 : p ≠ 0.9
2. The test is two tailed.
3. Reject the null hypothesis
Step-by-step explanation:
We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.
So, Null Hypothesis, [tex]H_0[/tex] : p = 0.90
Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.9
Here, the test is two tailed because we have given that to test the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.
Now, test statistics is given by;
[tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1) , where, n = sample size = 100
[tex]\hat p[/tex] = 0.94 (given)
So, Test statistics = [tex]\frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } }[/tex] = 1.68
Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)
= 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%
Now, our decision rule is that;
If p-value < significance level ⇒ Reject null hypothesis
If p-value > significance level ⇒ Accept null hypothesis
Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that proportion of people who own cats is significantly different than 90%.
A Type II error is defined as which of the following?
A. rejecting a false null hypothesis
B. rejecting a true null hypothesis
C. failing to reject a false null hypothesis
D. failing to reject a true null hypothesis
Answer:
Option C. failing to reject a false null hypothesis
Step-by-step explanation:
Type II error states that Probability of accepting null hypothesis given the fact that null hypothesis is false.
This is considered to be the most important error.
So, from the option given to us option C matches that failing to reject a false null hypothesis.
At St. Eligius Hospital, the weights of newborn babies follow a Normal distribution, with a mean of 7.5 pounds and a standard deviation of 1.2 pounds.
If Dr. Cavanero tells a mother that her newborn baby has a weight that is at the third quartile (or Q3), this means the baby must weigh approximately _____.
Answer:
The baby must weigh approximately 8.31 pounds.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 7.5 pounds
Standard Deviation, σ = 1.2 pounds
We are given that the distribution of weights of newborn babies is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the third quartile or [tex]Q_3[/tex]
We have to find the value of x such that the probability is 0.75
[tex]P( X < x) = P( z < \displaystyle\frac{x - 7.5}{1.2})=0.75[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 7.5}{1.2} = 0.674\\\\x = 8.3088 \approx 8.31[/tex]
Thus, the baby must weigh approximately 8.31 pounds.
The number of views of a page on a Web site follows aPoisson distribution with a mean of 1.5 per minute.a. What is the probability of no views in a minute?b. What is the probability of two or fewer views in10 minutes?c. Does the answer to the previous part depend on whetherthe 10-minute period is an uninterrupted interval? Explain.d. Determine the length of a time interval such that the probabilityof no views in an interval of this length is 0.001.
Answer:
a. P (X=0) = 0.223
b. P(X≤2) = 3.93 * 10^-5
C. No
D. t=4.6 miutes
Step-by-step explanation:
a. P (X=0) = [tex]e^{-1.5}[/tex] = 0.223
b. P(X≤2) =[tex]e^{1.5 * 10} (1 + 1.5*10 + \frac{1.5^{2} * 10^{2} }{2} )[/tex]
= 3.93 * 10^-5
C. No, the answer to the previous part does not depend on whether the 10-minute period is an uninterrupted interval, it only depend on the uniformity of the density of views per minute ad independency of the disjoint time intervals.
D. P (X=0) =0.001
[tex]e^{-1.5t} = 0.001\\\\-1.5t=ln(0.001)\\\\t =\frac{-6.9}{-1.5}[/tex]
t=4.6 miutes
The Poisson distribution is used to calculate the probabilities of certain counts of discrete events occurring in a given time interval. By using the average rate of occurrence, we can determine probabilities relevant to web traffic, call intervals, or urgent care visits.
Explanation:The Poisson distribution is used to model the number of events that occur in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. Various scenarios using Poisson distributions involve understanding the probability of a certain number of events occurring over a specified period.
Probability of no views in one minute: This can be computed using the formula for the Poisson distribution where the expected mean number of events (λ) is 1.5. The probability P(X=0) is given by (e^-1.5 * 1.5^0) / 0!.Probability of two or fewer views in 10 minutes: Given that events occur at a rate of 1.5 per minute, over 10 minutes the average rate would be 15. Calculate the cumulative probability for 0, 1, and 2 events and sum them up.Dependence of the interval: The answer to part b does not depend on the interval being uninterrupted, as the Poisson model assumes independence of intervals.Length of interval for a 0.001 probability of no views: Find the interval length 't' such that P(X=0) = 0.001 using the Poisson formula and solving for 't'.