Question 1

The ages of five randomly chosen cars in a parking garage are determined to be 7, 9, 3, 4, and 6 years old. If weconsider this sample of 5 in groups of 3, what is the probability of the population mean falling between 5.5 and6.5 years?
A) 0.5
B) 0.4
C) 0.55
D) 0.6

Accepted Answer

The answer of The ages of five randomly chosen cars...

Question 2

The amount of time it takes a student to walk from her home to class has a skewed right distribution with amean of 17 minutes and a standard deviation of 2.2 minutes. If data were collected from 50 randomly selectedwalks, describe the sampling distribution of $\overline { x _ { , } }$ the sample mean time.8.2 Distribution of the Sample Proportion1 Describe the sampling distribution of a sample proportion.

Accepted Answer

The answer of The amount of time it takes a...

Question 3

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standarddeviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability thatthe mean daily revenue for the next 30 days will exceed $7000?
A) 0.1814
B) 0.8186
C) 0.5675
D) 0.4325

Accepted Answer

The central limit theorem states that the distribution of sample means will be approximately normal, regardless of the shape of the population distribution, if the sample size is large enough. In this case, the sample size is 30, which is large enough to apply the central limit theorem.The mean of the sample means will be equal to the mean of the population, which is $7200. The standard deviation of the sample means will be equal to the standard deviation of the population divided by the square root of the sample size, which is $1200 / sqrt(30) = $219.09.The probability that the mean daily revenue for the next 30 days will exceed $7000 is the area under the normal curve to the right of $7000. This area can be calculated using a standard normal distribution table or a calculator. The area to the right of $7000 is 0.1814.Therefore, the correct answer is A) 0.1814.

Question 4

$$\mathrm { N } = 24,000 , \mathrm { n } = 400 , \mathrm { p } = 0.7$$
A) Approximately normal; $\mu _ { \mathrm { p } } = 0.7 , \sigma \mathrm { p } = 0.023$
B) Exactly normal; $\mu _ { \mathrm { p } } = 0.7 , \sigma _ { \mathrm { p } } = 0.023$
C) Approximately normal; $\mu _ { \mathrm { p } } = 0.7 , \sigma _ { \mathrm { p } } = 0.0935$
D) Binomial; $\mu _ { \mathrm { p } } = 280 , \sigma _ { \mathrm { p } } = 9.17$ Provide an appropriate response.

Accepted Answer

The distribution is approximately normal due to the large sample size (n = 400) and the use of the sample proportion (p = 0.7). The mean ($\mu_p$) of the sample proportion is indeed 0.7, and the standard deviation ($\sigma_p$) can be calculated using the formula for the standard deviation of a sample proportion: $\sigma_p = \sqrt{\frac{p(1-p)}{n}}$, which gives approximately 0.023 when p = 0.7 and n = 400.

Question 5

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standarddeviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability thatthe mean daily revenue for the next 30 days will be between $7000 and $7500?
A) 0.7333
B) 0.2667
C) 0.9147
D) 0.8186

Accepted Answer

The probability that the mean daily revenue for the next 30 days will be between $7000 and $7500 is 0.7333. This is because the mean daily revenue has a mean of $7200 and a standard deviation of $1200, and the probability that a random sample of 30 days will have a mean between $7000 and $7500 is given by the normal distribution.

Question 6

A farmer was interest in determining how many grasshoppers were in his field. He knows that the distributionof grasshoppers may not be normally distributed in his field due to growing conditions. As he drives histractor down each row he counts how many grasshoppers he sees flying away. After several rows he figuresthe mean number of flights to be 57 with a standard deviation of 12. What is the probability of the farmer willcount 52 or fewer flights on average in the next 40 rows down which he drives his tractor?
A) 0.0041
B) 0.4959
C) 0.9959
D) 0.0410

Accepted Answer

The answer of A farmer was interest in determining how...

Question 7

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time how many, how many groups can be formed?
A) 20
B) 10
C) 30
D) 15

Accepted Answer

There are 6 ways to choose the first bill, 5 ways to choose the second bill, and 4 ways to choose the third bill. So, there are 6 * 5 * 4 = 120 ways to choose 3 bills from 6 bills. However, since the order of the bills does not matter, we need to divide by 3! to account for the overcounting. So, there are 120 / 3! = 20 groups that can be formed.

Question 8

Professor Whata Guy surveyed a random sample of 420 statistics students. One of the questions was ʺWill youtake another mathematics class?ʺ The results showed that 252 of the students said yes. What is the sample^proportion, p of students who say they will take another math class?
A) 0.6
B) 0.42
C) 0.252
D) 0.775

Accepted Answer

The sample proportion $ p $ is calculated by dividing the number of students who said yes (252) by the total number of students surveyed (420), which gives $ \frac{252}{420} = 0.6 $.

Question 9

To assess attitudes towards issues that affect the residents of a village, the village randomly chose 800 familiesto participate in a survey of life attitudes. The village received 628 completed surveys. What is the sampleproportion of completed surveys?
A) 0.785
B) 0.628
C) 1.274
D) 0.886

Accepted Answer

The sample proportion of completed surveys is calculated by dividing the number of completed surveys (628) by the total number of surveys sent out (800), which equals 0.785.

Question 10

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time what is the probability of selecting a group that has an average value of equal to or less than $25?
A) 0.45
B) 0.86
C) 0.50
D) 0.35

Accepted Answer

The average value of the three bills is equal to or less than $25 if the sum of the values of the three bills is equal to or less than $75. There are 20 ways to select three bills from the six bills. The sum of the values of the three bills is equal to or less than $75 in 9 of the 20 ways. Therefore, the probability of selecting a group that has an average value of equal to or less than $25 is 9/20 = 0.45.

Question 11

A greenhouse in a tri-county area has kept track of its customers for the last several years and has determinedthat it has about 10,000 regular customers. Of those customers, 28% of them plant a vegetable garden in thespring. The greenhouse obtains a random sample of 800 of its customers. Is it safe to assume that the sampling^distribution of p, the sample proportion of customers that plant a vegetable garden, is approximately normal?
A) Yes
B) No

Accepted Answer

The sampling distribution of the sample proportion $p$ can be considered approximately normal if the sample size $n$ meets the conditions $np \geq 10$ and $n(1-p) \geq 10$. Given $n = 800$ and $p = 0.28$, $np = 800(0.28) = 224$ and $n(1-p) = 800(0.72) = 576$, both of which are greater than 10, thus satisfying the conditions for normal approximation.

Question 12

A farmer was interested in determining how many grasshoppers were in his field. He knows that thedistribution of grasshoppers may not be normally distributed in his field due to growing conditions. As hedrives his tractor down each row he counts how many grasshoppers he sees flying away. After several rows hefigures the mean number of flights to be 57 with a standard deviation of 12. What is the probability of thefarmer will count 52 or fewer flights or 60 or more flights on average in the next 40 rows down which he driveshis tractor?
A) 0.0612
B) 0.4959
C) 0.9388
D) 0.0530

Accepted Answer

The probability of counting 52 or fewer flights or 60 or more flights can be found using the normal distribution formula and Z-scores. Given a mean ($\mu$) of 57 and a standard deviation ($\sigma$) of 12 for the distribution of grasshopper flights, and considering the sample size (n) of 40, the standard error (SE) of the mean is $\sigma / \sqrt{n} = 12 / \sqrt{40}$. To find the Z-scores for 52 and 60, we use $Z = (X - \mu) / SE$. Calculating these Z-scores and looking up their corresponding probabilities in the Z-table, then adding the probabilities for Z-scores below 52 and above 60 gives us the total probability. The correct answer, based on these calculations, matches the probability given in option A, which accounts for the tails of the distribution where the farmer would see 52 or fewer and 60 or more flights.

Question 13

The ages of five randomly chosen cars in a parking garage are determined to be 7, 9, 3, 4, and 6 years old. If weconsider this sample of 5 in groups of 3, what is the probability of the population mean being more than 6years?
A) 0.4
B) 0.5
C) 0.1
D) 0.6

Accepted Answer

The answer of The ages of five randomly chosen cars...

Question 14

A national caterer determined that 87% of the people who sampled their food said that it was delicious. Arandom sample of 144 people is obtained from a population of 5000. The 144 people are asked to sample the^catererʹs food. Will the distribution of p, the sample proportion saying that the food is delicious, beapproximately normal?
A) Yes
B) No

Accepted Answer

The distribution of p will be approximately normal because the sample size is large enough (n=144) and both np and n(1-p) are greater than 10, satisfying the conditions for the Central Limit Theorem.

Question 15

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time what is the probability of selecting a group that has an average value of at least $26?
A) 0.55
B) 0.50
C) 0.60
D) 0.45

Accepted Answer

The answer of John has six bills of paper money...

Question 16

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time what is the probability of selecting a group that has an average value of between $40 and $45?
A) 0.15
B) 0.05
C) 0.25
D) 0.20

Accepted Answer

The only combination that has an average value between $40 and $45 is $20, $20, and $50. There are 20 possible combinations of 3 bills from 6, so the probability is 1/20 = 0.05.

Question 17

A national caterer determined that 87% of the people who sampled their food said that it was delicious. Arandom sample of 144 people is obtained from a population of 5000. The 144 people are asked to sample the^catererʹs food. If p is the sample proportion saying that the food is delicious, what is the mean of the sampling^distribution of p?
A) 0.87
B) 1.25
C) 0.19
D) 0.42

Accepted Answer

The mean of the sampling distribution of p (the sample proportion) is equal to the population proportion, which is given as 0.87.

Question 18

The ages of five randomly chosen cars in a parking garage are determined to be 7, 9, 3, 4, and 6 years old. If weconsider this sample of 5 in groups of 3, how many groups can be formed?
A) 10
B) 5
C) 30
D) 60

Accepted Answer

The number of groups of 3 that can be formed from 5 items is calculated using combinations: C(5,3) = 5! / (3!(5-3)!) = 10.

Question 19

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standarddeviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability thatthe mean daily revenue for the next 30 days will exceed $7500?
A) 0.0853
B) 0.9131
C) 0.0869
D) 0.9147

Accepted Answer

The probability that the mean daily revenue for the next 30 days will exceed $7500 is 0.0853. This is because the mean daily revenue is normally distributed with a mean of $7200 and a standard deviation of $1200. The z-score for a mean daily revenue of $7500 is (7500 - 7200) / 1200 = 0.25. The probability that the z-score will be greater than 0.25 is 0.0853.

Question 20

A national caterer determined that 37% of the people who sampled their food said that it was delicious. Arandom sample of 144 people is obtained from a population of 5000. The 144 people are asked to sample the^catererʹs food. If p is the sample proportion saying that the food is delicious, what is the standard deviation of^the sampling distribution of p?
A) 0.04
B) 0.002
C) 0.48
D) 0.23

Accepted Answer

The standard deviation of the sampling distribution of $p$ (the sample proportion) is calculated using the formula $\sqrt{\frac{p(1-p)}{n}}$, where $p = 0.37$ (the proportion of people who said the food was delicious) and $n = 144$ (the sample size). Plugging in these values, we get $\sqrt{\frac{0.37(1-0.37)}{144}} = \sqrt{\frac{0.37(0.63)}{144}} = \sqrt{\frac{0.2331}{144}} = \sqrt{0.00161875} \approx 0.04$.

Quiz 8: Sampling Distributions

The ages of five randomly chosen cars in a parking garage are determined to be 7, 9, 3, 4, and 6 years old. If weconsider this sample of 5 in groups of 3, what is the probability of the population mean falling between 5.5 and6.5 years?

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standarddeviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability thatthe mean daily revenue for the next 30 days will exceed $7000?

N=24,000,n=400,p=0.7\mathrm { N } = 24,000 , \mathrm { n } = 400 , \mathrm { p } = 0.7N=24,000,n=400,p=0.7

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standarddeviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability thatthe mean daily revenue for the next 30 days will be between $7000 and $7500?

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time how many, how many groups can be formed?

Professor Whata Guy surveyed a random sample of 420 statistics students. One of the questions was ʺWill youtake another mathematics class?ʺ The results showed that 252 of the students said yes. What is the sample^proportion, p of students who say they will take another math class?

To assess attitudes towards issues that affect the residents of a village, the village randomly chose 800 familiesto participate in a survey of life attitudes. The village received 628 completed surveys. What is the sampleproportion of completed surveys?

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time what is the probability of selecting a group that has an average value of equal to or less than $25?

The ages of five randomly chosen cars in a parking garage are determined to be 7, 9, 3, 4, and 6 years old. If weconsider this sample of 5 in groups of 3, what is the probability of the population mean being more than 6years?

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time what is the probability of selecting a group that has an average value of at least $26?

John has six bills of paper money in the following denominations: $1, $5, $10, $20, $50, and $100 If he selects 3bills at a time what is the probability of selecting a group that has an average value of between $40 and $45?

The ages of five randomly chosen cars in a parking garage are determined to be 7, 9, 3, 4, and 6 years old. If weconsider this sample of 5 in groups of 3, how many groups can be formed?

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standarddeviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability thatthe mean daily revenue for the next 30 days will exceed $7500?

$\mathrm { N } = 24,000 , \mathrm { n } = 400 , \mathrm { p } = 0.7$