Un'urna ha 2 palline bianche e 3 nere. Calcola la probabilità di estrazione della pallina bianca in 7 tentativi:
a. solo la prima volta;
b. una volta;
c. 5 volte;
d. sempre;
e. mai;
f. almeno una volta.
Calculate the estimated mean revision time for 120 students given the following time intervals and frequencies: 0<t≤15: 0, 15<t≤30: 20, 30<t≤45: 50, 45<t≤60: 50.
(a) Fill in the missing values in the frequency distribution table. (b) Find the mean: 113292≈2.6. (c) Determine the median. (d) Create a frequency histogram.
Find the central tendency measures (mode, median, mean) and compute range, IQR, quartile deviation, and standard deviation for the data: Marks: 0-20 (5), 20-40 (15), 40-60 (30), 60-80 (8), 80-100 (2).
Complete the sentence below.
The standard deviation of the sampling distribution of xˉ, denoted σxˉ, is called the of the . The standard deviation of the sampling distribution of xˉ, denoted σxˉ, is called the □□ of the □
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known Use the t-distribution table to find the critical value(s) for the indicated alternative hypotheses, level of significance α, and sample sizes n1 and n2. Assume that the samples are random and independent, and the populations are normally distributed. Complete parts (a) and (b).
Ha:μ1=μ2,α=0.20,n1=10,n2=2 Click the icon to view the t-distribution table.
(a) Find the critical value(s) assuming that the population variances are equal.
−1.372,1.372
(Type an integer or decimal rounded to three decimal places as needed. Use a comma to separate answers as needed.)
(b) Find the critical value(s) assuming that the population variances are not equal.
□ (Type an integer or decimal rounded to three decimal places as needed. Use a comma to separate answers as needed.)
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Save The accompanying data set includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are n=36,xˉ=12.192oz,s=0.099oz. Assume that a simple random sample has been selected. Use a 0.01 significance level to test the claim that cans of Coke have a mean volume of 12.00 ounces. Does it appear that consumers are being cheated? Click the icon to view the data set of regular Coke can volumes. Identify the null and alternative hypotheses.
H0 : □□□H1 : □□
(Type integers or decimals. Do not round.)
Use the t-distribution table to find the critical value(s) for the indicated alternative hypotheses, level of significance α, and sample sizes n1 and n2. Assume that the samples are random and independent, and the populations are normally distributed. Complete parts (a) and (b).
Ha:μ1<μ2,α=0.01,n1=15,n2=8 E
Click the icon to view the t-distribution table.
(a) Find the critical value(s) assuming that the population variances are equal.
□
(Type an integer or decimal rounded to three decimal places as needed. Use a comma to separate answers as needed.)
13. At a university, the probability that an incoming freshman will graduate within four years is 0.553 . What is the probability that at least 60 out of a group of 100 incoming freshman will graduate in four years?
For the following claim, find the null and alternative hypotheses, test statistic, critical value, and draw a conclusion. Assume that a simple random sample has been selected from a normally distributed population. Answer parts a-d. Claim: The mean IQ score of statistics professors is greater than 128. Sample data: n=17,xˉ=131,s=11. The significance level is α=0.05.
(7) Click the icon to view a table of critical t-values.
t=1.124 (Round to three decimal places as needed.)
c. Find the critical value using a t-distribution table. The critical value is □ (Round to three decimal places as needed.)
Question 13 of 20 Find the critical x2-values to test the claim σ2=4.3 if n=12 and α=0.05.
A. 3.053,24.725
B. 3.816,21.920
C. 4.575,19.675
D. 2.603,26.757
Question 15 of 20 Find the critical value and rejection region for the type of t-test with level of significance α and sample size n Left-tailed test, α=0.01,n=27
A. t0=−1.315;t<−1.315
B. t0=2.479;t>2.479
c. t0=−2.473;t<−2.473
D. t0=−2.479;t<−2.479
The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is skewed right. However, records indicate that the mean time is 11.7 minutes, and the standard deviation is 3.7 minutes. Complete parts (a) through (c) below. Click here to view the standard normal distribution table (page 1).
Click here to view the standard normal distribution table (page 2).
B. The sample size needs to be less than 30 .
C. The normal model cannot be used if the shape of the distribution is skewed right.
D. The sample size needs to be greater than 30 .
(b) What is the probability that a random sample of n=35 oil changes results in a sample mean time less than 10 minutes? The probability is approximately 0.0032 .
(Round to four decimal places as needed.)
(c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager. There would be a 10% chance of being at or below □ minutes.
(Round to one decimal place as needed.)
The most famous geyser in the world, Old Faithful in Yellowstone National Park, has a mean time between eruptions of 85 minutes. If the interval of time between the eruptions is normally distributed with standard deviation 21.25 minutes, complete parts (a) through (f). Click here to view the standard normal distribution table (page 1).
Click here to view the standard normal distribution table (page 2).
(a) What is the probability that a randomly selected time interval between eruptions is longer than 96 minutes? The probability that a randomly selected time interval is longer than 96 minutes is approximately 0.3025 .
(Round to four decimal places as needed.)
(b) What is the probability that a random sample of 6 time intervals between eruptions has a mean longer than 96 minutes? The probability that the mean of a random sample of 6 time intervals is more than 96 minutes is approximately 0.1024.
(Round to four decimal places as needed.)
(c) What is the probability that a random sample of 23 time intervals between eruptions has a mean longer than 96 minutes? The probability that the mean of a random sample of 23 time intervals is more than 96 minutes is approximately 0.0066 .
(Round to four decimal places as needed.)
(d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. If the sample size increases, the probability □ because the variability in the sample mean □
4.10. Ndryshorja e rastit X ka shpërndarje normale me pritje matematike μ=3 dhe dispersion σ2=9. Gjeni:
a) P(−2<X<5),P(X>0)
b) P(∣X−3∣>6),P(−2<X<−1)
c) P(−6<X<12),P(X<2)
d) P(0<X<1),P(X>2)
A local fast-food restaurant is running a "Draw a three, get it free" lunch promotion. After each customer orders, a touch-screen display shows the message, "Press here to win a free lunch." A computer program then simulates one card being drawn at random from a standard deck of playing cards. If the chosen card is a 3, the customer's order is free. (Note that the probability of drawing a 3 from a standard deck of playing cards is 4/52.) Otherwise, the customer must pay the bill. Suppose that 250 customers place lunch orders on the first day of the promotion. Let X= the number of people who win a free lunch.
Explain why X is a binomial random variable.
B- choose your answer... □ "failure"=Anything but a type your answer...
□ - "success"=Draw a type your answer...
□ - Knowing whether or not one person gets a type your answer...
□ tells you choose your answer...
□ about whether or not
1− choose your answer... □ another person gets a type your answer...
□
N- □−n= type your answer...
□ S- □p= type your answer...
□
6. (16 points) In a random survey of 900 licensed drivers aged 16 and over, 32% of those respondents reported that they make angry gestures while driving. Use a 0.05 significance level to test the claim that among licensed drivers aged 16 and over, the percentage who make angry gestures while driving is 30%. Identify the null hypothesis, alternative hypothesis, test statistic, critical value or P -value, and conclusion about the null hypothesis. Use the normal distribution as an approximation of the binomial distribution.
Determine whether the normal sampling distribution cain be used. The claim is p<0.015 and the sample size is n=150.
Use the normal distribution.
Do not use the normal distribution.
B= BINOMIAL PROBABILITY DISTRIBUTION FUNCTION Example 4.14
In the 2013 Jerry's Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X= the number of pages that feature signature artists.
What values does x take on?
What is the probability distribution? Find the following probabilities:
a) the probability that two pages feature signature artists
b) the probability that at most six pages feature signalure arists
c) the probability that more than three pages feature stonalure artists.
Using the formulas, calculate the (i) mean and (ii) slandard deviation.
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Suppose Z follows the standard normal distribution. Calculate the following probabilities using the ALEKS calculator. Round your responses to at least three decimal places.
(a) P(z≤−1.18)=□
(b) P(z>0.77)=□
(c) P(−0.65<Z<2.02)=□
\begin{tabular}{|l|l|}
\hline What are the chances of them adding up to 6 or 8 ? & 0.167 \\
\hline What are the chances of them adding up to a prime number? & \\
\hline What are the chances of rolling a double (same number on each die)? \\
\hline What are the chances of rolling snake-eyes twice in a row? & \\
\hline
\end{tabular}
A rainstorm in Portland, Oregon, wiped out the electricity in 5% of the households in the city. Suppose that a random sample of 50 Portland households is taken after the rainstorm. Answer the following.
(If necessary, consult a list of formulas.)
(a) Estimate the number of households in the sample that lost electricity by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.
□
(b) Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.
□
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1 point A survey found that 25% of pet owners have their pets bathed professionally than doing it themselves. If 18 pet owners are surveyed, what is the probability that no more than 6 pet owners have their pets professionally bathed? (Round answer to the third decimal) Type your answer...
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A test has 10 multiple choice questions each of which has four choices and only one right answer. A student decides to use a spinner with four equivalent sectors on it to randomly choose which answer to pick for each question. What is the likelihood the student gets no more than 2 of the questions correct? (Round to the third decimal)
A test has 10 multiple choice questions each of which has four choices and only one right answer. A student decides to use a spinner with four equivalent sectors on it to randomly choose which answer to pick for each question. What is the likelihood the student gets between 7 and 9 (inclusive) of the questions correct?
(Round to the fourth decimal)
In a binomial distribution if it is expected that 56% of people will have pets then when a sample of 125 people are selected it is expected that 70 will not have pets.
True
False
A chef is trying out a new recipe. From his past experience, it takes him a few trials to get it right. He estimates for the first 15 trials, his probability of getting the recipe right is 35% in each trial. Since it's new, every trial is considered to be independent from the other. Submit all your answers to three decimal places.
(a) The chef would like to try the recipe 10 times, and he wants to add it to the menu if he gets at least 6 right.
(i) What is the probability that the new recipe will be added to the menu?
0.021
(ii) What is the probability that the chef gets exactly 6 right?
(b) The chef would like to try the recipe 15 times.
(i) What is the probability that the chef gets exactly 6 right?
Let × be a discrete random variable with the following PMF. Answer questions 1,2 and 3
Px(x)=⎩⎨⎧k1k18181k10 for x=−2 for x=−1 for x=0 for x=1 for x=2 otherwise 1. (1 point) Find the value of k
A. 0.125
B. 0.25
C. 8
D. 1
E. 4 2. (1 point) Find P(−1.5<x<0.5)
A. 0.25
B. 1
C. 0.375
D. 5/32
E. 0.75 3. (1 point) The E(X) equals:
A. 1
B. 2
C. 0
D. 0.125
E. -0.125 The discrete random variable X takes the values 1,2 and 3 and has cumulative distribution function F(x) given by
\begin{tabular}{|c|c|c|c|}
\hline X & 1 & 2 & 3 \\
\hline F(x) & 0.4 & 0.4 & 1 \\
\hline
\end{tabular} 4. (2 points) The variance of X equals:
A. 1.7
B. 0.96
C. 0.91
D. 0.98
E. 5.8
Page 217
A fine dining restaurant claims that the group sizes of their customers follows the following distribution:
\begin{tabular}{|c|c|c|c}
\hline 2 people & 3 people & 4 people & 4+ people \\
\hline 36% & 10% & 17% & 37%
\end{tabular} Gustavo, a waiter at the restaurant, would like to test this claim. He takes a sample of 93 customers and records the observed frequencies in the following table:
\begin{tabular}{|c|c|c|c}
2 people & 3 people & 4 people & 4+ people \\
\hline 35 & 2 & 21 & 35
\end{tabular}
(a) In performing this statistical test, state the hypotheses.
- Ho: the distribution of customers for each group is the same as the observed frequencies vs. Ha: the distribution of customers for each group is not the same as the observed frequencies
Ho: the distribution of customers for each group is not the same as claimed by the restaurant vs. Ha: the distribution of customers for each group is the same as claimed by the restaurant
Ho: the distribution of customers for each group is the same as claimed by the restaurant vs. Ha: the distribution of customers for each group is not the same as claimed by the restaurant
Ho: the distribution of customers for each group is not the same as the observed frequencies vs. Ha: the distribution of customers for each group is the same as the observed frequencies
Ho: the proportions of customers for each group are all the same vs. Ha: the proportions of customers for each group are not all the same
\begin{tabular}{|c|c|}
\hline What are the chances of them adding up to 7 ? & 0.167 \\
\hline What are the chances of them adding up to 6 or 8? & 0.42 \\
\hline What are the chances of them adding up to a prime number? & 0.42 \\
\hline What are the chances of rolling a double (same number on each die)? & 0.167 \\
\hline What are the chances of rolling snake-eyes twice in a row? & 0.42 \\
\hline What are the chances of the dice adding up to an even number? & 0.5 \\
\hline What are the chances that both dice will be even numbers? & 0 \\
\hline
\end{tabular}
The Health Department states the mean pregnancy term is 8.60 months. You conduct a study of 35 women in the Reedley area and find the mean pregnancy term is 8.68 months and a sample standard deviation of 0.2 months. Using a .01 level of significance, can you conclude that the mean pregnancy term in Reedley is different than 8.6 months? Be sure to include all of the sections listed below. Use the picture provided to choose how your curve should look and be labeled. State H0 and H1.
Use the picture and tell me the letter of the curve that should be used.
State your critical value(s).
Find the test statistic.
Find the mode(s) for the data items in the given frequency distribution.
\begin{tabular}{|l|c|c|c|c|c|c|c|c|}
\hline Score, x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline Frequency, f & 2 & 5 & 1 & 4 & 4 & 4 & 1 & 6 \\
\hline
\end{tabular} Select the correct choice below and, if necessary, fill in the answer box within your choice.
A. The mode(s) is/are □ (Use a comma to separate answers as needed.)
B. There is no mode.
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Home - Northern Essex Community College... Homework : 4:7(1,2,3,4)8(2,3,4)
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(a) Find the probability that the waiting time is less than 4 minutes. The probability that the waiting time is less than 4 minutes is □ .
Homework ; 4:7(1,2,3,4)8(2,3,4)
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Question 1 of 40 (1 point) I Question Attempt: 1 of 3
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Part 1 of 3
(a) Find the probability that the waiting time is less than 4 minutes. The probability that the waiting time is less than 4 minutes is 52. Part 2 of 3
(b) Find the probability that the waiting time is greater than 3 minutes. The probability that the waiting time is greater than 3 minutes is 107. Part: 2/3 Part 3 of 3
(c) Find the probability that the waiting time is between 2 and 8 minutes. The probability that the waiting time is between 2 and 8 minutes is □.
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Homework \# 4: 7(1,2,3,4) 8(2,3,4)
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11 The following figure is a probability density curve that represents the grade point averages (GPA) of the graduating seniors at a large university. Part 1 of 2 Find the proportion of seniors whose GPA is between 3.1 and 3.4.
The proportion of seniors whose GPA is between 3.1 and 3.4 is □ . Part 2 of 2 What is the probability that a randomly chosen senior will have a GPA greater than 3.4 ?
The probability that a randomly chosen senior will have a GPA greater than 3.4 is □ .
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Al surveyed 60 people on restaurant spending. a) Identify the modal class. b) Estimate the mean. c) Estimate the standard deviation and discuss it. d) Estimate the variance, range, and interquartile range, explaining why they are estimates.
Find the grouped frequency distribution for these test scores: 83, 85, 97, 91, 92, 82, 90, 89, 91, 83, 93, 88, 86, 84, 98. Class width is 5. Then, draw a frequency polygon with midpoints labeled.
Micro-Pub, Inc. is evaluating two cameras (R and S). a. Find the rate of return range for both cameras.
b. Calculate the expected return for each camera.
c. Which camera is riskier and why? Initial investment: \$4,000 for both.
Camera R: Pessimistic 20%, Most likely 25% (0.50), Optimistic 30% (0.25).
Camera S: Pessimistic 15% (0.20), Most likely 25% (0.55), Optimistic 35% (0.25).
Homework \# 4: 7(1,2,3,4)8(2,3,4)
Question 7 of 40 (1 point) | Question Attempt: 1 of 3
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9 A normal distribution has mean μ=61 and standard deviation σ=20. Find and interpret the z-score for x=63. The z-score for x=63 is □ . So 63 is □ standard deviations (Choose one) the mean μ=61.
pago 34 Solution. (a) On a
fX,Y(x,y)={10xy2,0≤x≤y≤10, ailleurs Pour 0<x<y<1, on a FX,Y(x,y)=∫0x∫uy10uv2dvdu=10∫0xu∫uyv2dvdu=10∫0xu(3v3)∣∣uydu=10∫0xu(3y3−u3)du=310∫0xy3u−u4du=310(y32u2−5u5)∣∣0x=310(y32x2−5x5)=35x2y3−32x5
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Homework * 4: 7(1,2,3,4)8(2,3,4)
Question 8 of 40 (1 point) I Question Attempt 1 of 3
Jonathan
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12 A normal population has mean μ=37 and standard deviation σ=14. Find the value that has 25% of the population above it. Round the answer to at least one decimal place. The value that has 25% of the population above it is □ .
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Homework \# 4: 7(1,2,3,4)8(2,3,4)
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✓7✓✓9 A normal population has mean μ=9 and standard deviation σ=5.
(a) What proportion of the population is less than 20?
(b) What is the probability that a randomly chosen value will be greater than 6 ? Round the answers to four decimal places. Part: 0/2 Part 1 of 2 The proportion of the population less than 20 is □ .
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Homework \# 4: 7(1,2,3,4) 8(2,3,4)
Question 11 of 40 (1 point) I Question Attempt: 1 of 3
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(a) What proportion of the population is less than 20?
(b) What is the probability that a randomly chosen value will be greater than 6 ? Round the answers to four decimal places. Part 1 of 2 The proportion of the population less than 20 is 0.9861 . Part: 1 / 2 Part 2 of 2 The probability that a randomly chosen value will be greater than 6 is □ .
1
Matching
5 points Determine if each of the following situations is a binomial distribution. If it does, explain why each condition is met. If it does not, say what condition was not met I roll a die 100 times and count how many of each number I get. I flip a coin and count how many times it takes to land on a "head". I spin a spinner 20 times and count how many times it lands on blue. I flip a coin 12 times and count how many heads I get.
I throw a bean bag at a cornhole board 10 times. I receive coaching after each throw. I count how many bags go in.
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✓11 A sample of size 190 will be drawn from a population with mean 45 and standard deviation 11. Use the TI-83 Plus/TI-84 Plus calculator.
Part 1 of 2
(a) Find the probability that xˉ will be less than 43 . Round the answer to at least four decimal places. The probability that xˉ will be less than 43 is 0.0060 . Part: 1/2 Part 2 of 2
(b) Find the 60th percentile of xˉ. Round the answer to at least two decimal places. The 60th percentile is □.
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Español Watch your cholesterol: The mean serum cholesterol level for U.S. adults was 198, with a standard deviation of 39.3 (the units are milligrams per deciliter). A simple random sample of 106 adults is chosen. Use the TI-84 Plus calculator. Round the answers to at least four decimal places. Part 1 of 3
(a) What is the probability that the sample mean cholesterol level is greater than 205? The probability that the sample mean cholesterol level is greater than 205 is 0.0331 . Part 2 of 3
(b) What is the probability that the sample mean cholesterol level is between 186 and 192? The probability that the sample mean cholesterol level is between 186 and 192 is 0.0572 . Part: 2/3 Part 3 of 3
(c) Using a cutoff of 0.05 , would it be unusual for the sample mean to be less than 194? It □ (Choose one) be unusual for the sample mean to be less than 194, since the probability is □ .
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Jonathan TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household was 2.24. Assume the standard deviation is 1.2 . A sample of 95 households is drawn. Part: 0/5□
Part 1 of 5
(a) What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to at least four decimal places. The probability that the sample mean number of TV sets is greater than 2 is □ .
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Español TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household was 2.24. Assume the standard deviation is 1.2 . A sample of 95 households is drawn. Part 1 of 5
(a) What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to at least four decimal places. The probability that the sample mean number of TV sets is greater than 2 is 0.9744 . Part: 1/5 Part 2 of 5
(b) What is the probability that the sample mean number of TV sets is between 2.5 and 3? Round your answer to at least four decimal places. The probability that the sample mean number of TV sets is between 2.5 and 3 is □ .
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There are 48 states in a certain country. The histogram shows murder rates per 100,000 residents and the number of states that had these rates in a certain year. Use this information to complete parts (a) through (e) to the right. Murder Rates per 100,000 Residents, by State
a. Is the shape of this distribution best classified as normal, skewed left, or skewed right? Choose the correct distribution type below.
A. Normal distribution
B. Skewed left
c. Skewed right
b. Uniform distribution
b. Calculate the mean murder rate per 100,000 residents for the states. The mean murder rate is □ murders per 100,000 residents. (Type an integer or decimal rounded to two decimal place as needed.)
The scores on a test are normally distributed with a mean of 200 and a standard deviation of 10 . Find the score that is 321 standard deviations above the mean. A score of □ is 321 standard deviations above the mean.
This question: 1
point(s) possible
Subm Find the mean for the data items in the given frequency distribution.
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline Score, x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline Frequency, f & 3 & 6 & 2 & 5 & 3 & 1 & 3 & 2 \\
\hline
\end{tabular} The mean is □ (Round to 3 decimal places as needed.)
Find the critical values for a 95% confidence interval using the chi-square distribution with 20 degrees of freedom. Round the answers to three decimal places. The critical values are □ and □ .
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20 Coffee: The National Coffee Association reported that 65% of U.S. adults drink coffee daily. A random sample of 300U.S. adults is selected. Round your answers to at least four decimal places as needed. Part 1 of 6
(a) Find the mean μp. The mean μp^ is 0.65 . Part 2 of 6
(b) Find the standard deviation σp^. The standard deviation σp^ is 0.0275 . Part 3 of 6
(c) Find the probability that more than 66% of the sampled adults drink coffee daily. The probability that more than 66% of the sampled adults drink coffee daily is 0.3564 . Part 4 of 6
(d) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 . The probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 is 0.9836 .
Part 5 of 6
Cava For 1 ater
Submi
(e) Find the probability that less than 60% of sampled adults drink coffee daily. The probability that less than 60% of sampled aduts drink coffee đaity is □ .
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13 Coffee: The National Coffee Association reported that 65% of U.S. adults drink coffee daily. A random sample of 300 U.S. adults is selected. Round your answers to at least four decimal places as needed. Part 1 of 6
(a) Find the mean μp. The mean μp^ is 0.65 . Part 2 of 6
(b) Find the standard deviation σp^. The standard deviation σp^ is 0.0275 . Part 3 of 6
(c) Find the probability that more than 66% of the sampled adults drink coffee daily. The probability that more than 66% of the sampled adults drink coffee daily is 0.3564 . Part 4 of 6
(d) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 . The probability that the proportion of the sampled adults who drink coffee daily is between 0.57 and 0.71 is 0.9836 .
Part: 5/6 Part 6 of 6
(f) Using a cutoff of 0.05 , would it be unusual if less than 63% of the sampled adults drink coffee daily?
it □ (Choose one) V be unusual if less than 63% of the sampled adults drink coffee daily, since the probability is □ would would not
Coffee: The National Coffee Association reported that 63% of U.S. adults drink coffee daily. A random sample of 250 U.S. adults is selected. Round your answers to at least four decimal places as needed. Part 1 of 2
(a) Find the probability that more than 64% of the sampled adults drink coffee daily. The probability that more than 64% of the sampled adults drink coffee daily is 0.3721 . Part: 1/2 Part 2 of 2
(b) Find the probability that the proportion of the sampled adults who drink coffee daily is between 0.62 and 0.68 . The probability that the proportion of the sampled adults who drink coffee daily is between 0.62 and 0.68 is □.
Homework \# 4: 7(1,2,3,4)8(2,3,4)
Question 21 of 40 (1 point) I Question Attempt: 1 of 3
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≡21✓22 Student loans: The Institute for College Access and Success reported that 65% of college students in a recent year graduated with student loan debt. A random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary. Part 1 of 6
(a) Find the mean μp^. The mean μp is 0.65 . Part: 1 / 6 Part 2 of 6
(b) Find the standard deviation σp^. The standard deviation σp^ is □
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\text{Student loans: The Institute for College Access and Success reported that } 65\% \text{ of college students in a recent year graduated with student loan random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary.} \text{Part 1 of 6}
\begin{itemize}
\item[(a)] \text{Find the mean } \mu_{\hat{p}}.
\begin{itemize}
\item \text{The mean } \mu_{\hat{p}} \text{ is } 0.65.
\end{itemize}
\end{itemize} \text{Part 2 of 6}
\begin{itemize}
\item[(b)] \text{Find the standard deviation } \sigma.
\begin{itemize}
\item \text{The standard deviation } \sigma_{\hat{p}} \text{ is } 0.0503.
\end{itemize}
\end{itemize} \text{Part 3 of 6}
\begin{itemize}
\item[(c)] \text{Find the probability that less than } 52\% \text{ of the people in the sample were in debt.}
\begin{itemize}
\item \text{The probability that less than } 52\% \text{ of the people in the sample were in debt is } 0.0049.
\end{itemize}
\end{itemize} \text{Part 4 of 6}
\begin{itemize}
\item[(d)] \text{Find the probability that between } 60\% \text{ and } 80\% \text{ of the people in the sample were in debt.}
\begin{itemize}
\item \text{The probability that between } 60\% \text{ and } 80\% \text{ of the people in the sample were in debt is } 0.8375.
\end{itemize}
\end{itemize} \text{Part 5 of 6}
\begin{itemize}
\item[(e)] \text{Find the probability that more than } 70\% \text{ of the people in the sample were in debt.}
\begin{itemize}
\item \text{The probability that more than } 70\% \text{ of the people in the sample were in debt is } \square.
\end{itemize}
\end{itemize}
\text{Student loans: The Institute for College Access and Success reported that } 65\% \text{ of college students in a recent year graduated with student loan debt. A random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary.} \text{Part 1 of 6}
\text{(a) Find the mean } \mu_{\hat{p}}. \text{The mean } \mu_{\hat{p}} \text{ is } 0.65. \text{Part 2 of 6}
\text{(b) Find the standard deviation } \sigma_{\hat{p}}. \text{Part 6 of 6}
\text{(f) Using a cutoff of } 0.05, \text{ would it be unusual if less than } 66\% \text{ of people in the sample were in debt?} \text{It (Choose one) } \nabla \text{ be unusual if less than } 66\% \text{ of the people in the sample were in debt, since the probability is } \square.
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25> SAT scores: Assume that in a given year the mean mathematics SAT score was 522, and the standard deviation was 116 . A sample of 66 scores is chosen. Use the TI-84 Plus calculator. Part 1 of 5
(a) What is the probability that the sample mean score is less than 509? Round the answer to at least four decimal places. The probability that the sample mean score is less than 509 is 0.1841R◯. Correct Answer: The probability that the sample mean score is less than 509 is 0.1813 . Part: 1/5 Part 2 of 5
(b) What is the probability that the sample mean score is between 486 and 525? Round the answer to at least four decimal places. The probability that the sample mean score is between 486 and 525 is □
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Question 16 of 40 (1 point) I Question Attempt: 1 of 3
Jonathan SAT scores: Assume that in a given year the mean mathematics SAT score was 522, and the standard deviation was 116. A sample of 66 scores is chosen. Use the TI-84 Plus calculator.
Español Part 1 of 5
(a) What is the probability that the sample mean score is less than 509? Round the answer to at least four decimal places. The probability that the sample mean score is less than 509 is 0.18418. Correct Answer: The probability that the sample mean score is less than 509 is 0.1813 . Part 2 of 5
(b) What is the probability that the sample mean score is between 486 and 525? Round the answer to at least four decimal places. The probability that the sample mean score is between 486 and 525 is 0.5773 Part: 2/5 Part 3 of 5
(c) Find the 90th percentile of the sample mean. Round the answer to at least two decimal places. The 90th percentile of the samplemean is □ .
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Question 16 of 40 (1 point) I Question Attempt 1 of 3
Jonathan The probability that the sample mean score is less than 509 is 0.1841 0
Españ Correct Answer: The probability that the sample mean score is less than 509 is 0.1813 . Part 2 of 5
(b) What is the probability that the sample mean score is between 486 and 525? Round the answer to at least four decimal places. The probability that the sample mean score is between 486 and 525 is 0.5773 Part 3 of 5
(c) Find the 90th percentile of the sample mean. Round the answer to at least two decimal places. The 90th percentile of the sample mean is 540.25 . Part: 3/5 Part 4 of 5
(d) Using a cutoff of 0.05 , would it be unusual if the sample mean were greater than 525 ? Round the answer to at least four decimal places. It □ (Choose one) be unusual if the sample mean were greater than 525 , since the probability is □ .
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25 Taxes: The Internal Revenue Service reports that the mean federal income tax paid in the year 2010 was $8040. Assume that-the standard deviation is $4500. The IRS plans to draw a sample of 1000 tax returns to study the effect of a new tax law. Part 1 of 5
(a) What is the probability that the sample mean tax is less than $7900 ? Round the answer to at least four decimal places. The probability that the sample mean tax is less than $7900 is 0.1628 Part: 1/5 Part 2 of 5
(b) What is the probability that the sample mean tax is between $7400 and $8100 ? Round the answer to at least four decimal places. The probability that the sample mean tax is between $7400 and $8100 is 0.9986$.
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Jonatha Taxes: The Internal Revenue Service reports that the mean federal income tax paid in the year 2010 was $8040. Assume that the standard deviation is $4500. The IRS plans to draw a sample of 1000 tax returns to study the effect of a new tax law. Part 1 of 5
(a) What is the probability that the sample mean tax is less than $7900 ? Round the answer to at least four decimal places. The probability that the sample mean tax is less than $7900 is 0.1628 . Part 2 of 5
(b) What is the probability that the sample mean tax is between $7400 and $8100 ? Round the answer to at least four decimal places. The probability that the sample mean tax is between $7400 and $8100 is 0.6628 .
Correct Answer: The probability that the sample mean tax is between $7400 and $8100 is 0.6634 . Part: 2/5 Part 3 of 5
(c) Find the 80th percentile of the sample mean. Round the answer to at least two decimal places. The 80th percentile of the sample mean is $□
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iven in the table are the BMI statistics for random samples of men and women. Assume that the two samples are independent simple random samples selected from normally istributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts.
Submit t
InIS test: 20 point(S) possible
This question: 4 point(s) possible
\begin{tabular}{|c|c|r|}
\hline & Male BMI & Femal \\
\hline μ & μ1 & μ2 \\
\hline nˉ & 49 & 49 \\
\hline xˉ & 27.8756 & 26.07 \\
\hline s & 8.866451 & 4.1461 \\
\hline
\end{tabular}
- Test the claim that males and females have the same mean body mass index (BMI). That are the null and alternative hypotheses?
A
H0:μ1=μ2H1:μ1=μ2
B. H0:μ1=μ2H1:μ1<μ2
C
H0:μ1=μ2H1:μ1>μ2
D. H0:μ1≥μ2H1:μ1<μ2
he test statistic, t , is □ . (Round to two decimal places as needed.)
he P-value is □ (Round to three decimal places as needed.)
tate the conclusion for the test.
A. Reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.
B. Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.
C. Fail to reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI.
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Using the following data answer questions 6 \& 7 :
\begin{tabular}{|c|l|}
\hline Value & Frequency \\
\hline 1 & X \\
\hline 2 & XX \\
\hline 3 & XXX \\
\hline 4 & XXXX \\
\hline 5 & XXXXX \\
\hline 6 & X \\
\hline
\end{tabular}
histogram A HISTOGRAM
B HISTOGRAM C
histogRam D 6. (1 point) Which histogram represents the data? 7. (1 point) What is the skewness of the data?
a. Left skew
b. Right skew
c. Bimodal
d. Symmetric 8. (1 point) The standard deviation of −5,−5,−5,−5,−5 is
a. -5
b. 5
c. 0
d. -25 9. (1 point) The mean of 11 numbers is 7 . One of the numbers, 13 , is deleted. What is the mean of the remaining 10 numbers?
a. 7.7
b. 6.4
c. 6.0
The incomes in a certain large population of college teachers have a Normal distribution, with mean \75,000andstandarddeviation\10,000. Sixteen teachers are selected at random from this population to serve on a committee. What is the probability that their average salary is more than $77,500 ?
0.8413
essentially 0
0.0228
0.1587
Q1(6). Parents were asked if sports are equally important for boys and girls. Suppose that the proportion of parents who agree the equal importance in the population is actually 0.65 . In a survey, of the 400 parents interviewed, 70% agreed that boys and girls should have equal opportunities to participate in sports. 1(2). Describe the sampling distribution of the sample proportion of parents who agree that boys and girls should have equal opportunities. 2(4). What is the probability of observing a sample proportion as large as or larger than the observed value p^=0.70?
Note: Figure not drawn to scale. A long insulating tube of length L has inner radius ra and outer radius rb, where L≫rb, as shown in the figure. The tube has a nonuniform charge distributed throughout its volume. The charge density ρ as a function of the distance r from the tube's central axis is ρ(r)=βr3 for ra<r<rb, where β is a positive constant. lgnoring edge effects, what is the electric field E in the range ra<r<rb inside the tube?
(A) 0
(B) 4εarβ(r4−ra4)
(C) 5ε0rβ(r5−ra5)
(D) 5ε0rβ(rb5−r5)
Question 6 of 24
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The gates of 80 students have been recorded showing man of 18 and variance of 16 . Then the proportion of thee grades will tall betweren 8 and 78 is
At least 56\%
At least 89\%
Approximately 18%
Approximately 99.7\%
At least 25\%
At least 84\%
Not yet answered
Marked out of 1.00 The grades of 50 students have been recorded showing mean of 12 and variance of 9 . Note that the median mode, and the mean are equal. Then the number of these grades will fall between 3 and 12 is approximatehy
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How can we use the distributive property to find an expression equivalent to 4(5−3x) ? Place the steps in order.
CLEAR
CHECK Subtract the products.
Multiply 4 by 5 and multiply 4 by 3x. Subtract 3x from 5 to get 2x.
Multiply 4 by 2x. Step 1
DRAG AND DROP
Step 1⎩⎨⎧ DRAG AND DROP AN ITEM HERE Step 2 DRAG AND DROP AN ITEM HERE
Here are the hottest recorded temperatures (in ∘F ) for each of sixteen cities throughout North America.
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline \multicolumn{7}{|c|}{ Temperatures } \\
( in ∘F)
\end{tabular} Send data to calculator
(a) Complete the grouped frequency distribution for the data. (Note that the class width is 4. )
\begin{tabular}{|cc|}
\hline \begin{tabular}{c}
Temperatures \\
(in ∘F )
\end{tabular} & \begin{tabular}{c}
Frequency
\end{tabular} \\
\hline 92.5 to 96.5 & □ \\
96.5 to 100.5 & □ \\
100.5 to 104.5 & □ \\
104.5 to 108.5 & □ \\
108.5 to 112.5 & □ \\
\hline
\end{tabular}
(b) Using the classes from part (a), draw the frequency polygon for the data. Note that you can add or remove classes from the figure. Label each class with its midpoint.
\begin{tabular}{|c|c|c|c|c|}
\hline \multicolumn{5}{|c|}{ Test scores } \\
\hline 97 & 80 & 94 & 86 & 85 \\
\hline 96 & 80 & 83 & 84 & 95 \\
\hline 98 & 93 & 94 & 81 & 82 \\
\hline
\end{tabular} Send data to calculator
(a) Complete the grouped frequency distribution for the data. (Note that the class width is 5.)
\begin{tabular}{|cc|}
\hline Test scores & Frequency \\
\hline 78.5 to 83.5 & 5 \\
83.5 to 88.5 & 3 \\
88.5 to 93.5 & 1 \\
93.5 to 98.5 & 6 \\
\hline
\end{tabular}
(b) Using the classes from part (a), draw the frequency polygon for the data. Note that you can add or remove classes from the figure. Label each class with its midpoint.
Huntington-Hill method The fire department of a certain city is adding 65 firefighters to its force. The fire chief needs to allocate the firefighters to five districts, according to their populations. The populations of the districts are shown in the table below. Use the Huntington-Hill method to allocate the firefighters to the districts.
You may use the apportionment tool to help you. To do this, turn ON the apportionment tool, enter a divisor, and click on "Compute". You will then see the quota, the lower quota, the rounded quota (to the nearest whole number), the upper quota, and the geometric mean of the lower quota and upper quota for each district. You may also click on active cells to fill in the bottom row. Apportionment Tool:
OFF)
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline District & Natal & Draco & Salto & York & Tabriz & Total \\
\hline Population & 65,380 & 130,695 & 94,890 & 43,440 & 47,145 & 381,550 \\
\hline \begin{tabular}{c}
Number of \\
firefighters
\end{tabular} & □ & □ & □ & □ & □ & 0 \\
\hline
\end{tabular}
Find the standard deviation of human pregnancy lengths given a mean of 267 days and 95% lasting between 245 and 289 days. What percent last at least 285 days? Standard deviation: □ days.
Swift Manufacturing evaluates two projects. Find the return range, expected return, standard deviation, and coefficient of variation for each project. Also, create bar charts for the returns and determine which project is less risky.