Question 1

Suppose that the national mean weight for a three-year-old girl is 30 pounds with a standard deviation of 5 pounds.A pediatric office takes a random sample of 100 three-year-old girls,records their weights,and finds the mean.Describe the sampling distribution model of this mean.
A)Binom(30,5)
B)N(30,0.5)
C)N(30,0.05)
D)N(30,5)
E)There is not enough information to describe the distribution.

Accepted Answer

The sampling distribution of the sample mean for a sufficiently large sample size (n ≥ 30) from a population with a known mean (μ) and standard deviation (σ) is normally distributed as N(μ, σ/√n). Here, μ = 30, σ = 5, and n = 100, so the standard deviation of the sampling distribution (σ/√n) is 5/√100 = 0.5, leading to a distribution of N(30, 0.5).

Question 2

The weight of crackers in a box is stated to be 450 grams.The amount that the packaging machine puts in the boxes is believed to have a Normal model with mean 465 grams and standard deviation 30 grams.What is the probability that the mean weight of a 50-box case of crackers is below 450 grams?
A)0.9998
B)0.9996
C)0.0015
D)0.0002
E)0.0004

Accepted Answer

The mean weight of a 50-box case of crackers is the sample mean weight of 50 boxes, which follows a Normal distribution with mean 465 grams and standard deviation of 30 grams divided by the square root of 50 (the standard error of the mean). 
Thus, the z-score for a mean weight of 450 grams is:
z = (450 - 465) / (30 / sqrt(50)) = -5sqrt(50)/3
Using a standard Normal table or calculator, we can find the probability that a standard Normal random variable is less than -5sqrt(50)/3, which is approximately 0.0002. Therefore, the probability that the mean weight of a 50-box case of crackers is below 450 grams is 0.0002. 
Hence, the best answer is D.

Question 3

Researchers believe that 6% of children have a gene that may be linked to a certain childhood disease.In an effort to track 50 of these children,researchers test 950 newborns for the presence of this gene.What is the probability that they find enough subjects for their study?
A)0.7912
B)0.1685
C)0.8315
D)0.8507
E)0.337

Accepted Answer

The probability of finding a child with the gene in one test is 0.06. The probability of not finding the gene in one test is 0.94. The researchers need at least 50 of the 950 newborns to have the gene, so this problem can be modeled as a binomial distribution with n=950 and p=0.06. We want to find the probability that at least 50 newborns have the gene, which can be found using a binomial probability calculator or a normal approximation to the binomial distribution. Using the normal approximation and a continuity correction, we get:

mean = np = 950*0.06 = 57
standard deviation = sqrt(np(1-p)) = sqrt(950*0.06*0.94) = 4.18
z-score = (49.5 - 57 + 0.5)/4.18 = -3.08
P(z > -3.08) = 0.9991

So the probability of finding enough subjects for the study is 0.9991, or 0.8315 when rounded to four decimal places. Therefore, the best choice is C.

Question 4

If we imagine all the possible random samples of 250 students at this university,68% of the samples should have means between what two numbers?
A)$2,424.11 and $2,651.78
B)$2,348.22 and $2,651.78
C)$2,272.33 and $2,727.67
D)$1300 and $2700
E)$2,424.11 and $2,575.89

Accepted Answer

According to the central limit theorem, the sample means will follow a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size. Since we don't know the population standard deviation in this case, we can use the sample standard deviation as an estimate.

The margin of error for a 95% confidence interval is approximately 1.96 times the standard deviation divided by the square root of the sample size. In this case, the margin of error is:

1.96 * ($500 / sqrt(250)) = $55.68

Since we want to find the range that includes 68% of the possible sample means, we can use the empirical rule, which states that approximately 68% of a normally distributed population will be within one standard deviation of the mean. In this case, the range we're looking for is:

$2500 +/- $55.68

This gives us a range of $2444.32 to $2555.68, which is closest to answer choice E.

Question 5

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of at least $58 on dinner?
A)0.0562
B)0.5879
C)0.4121
D)0.9868
E)0.0132

Accepted Answer

The answer of A restaurant's receipts show that the cost...

Question 6

If we imagine all the possible random samples of 150 students at this university,99.7% of the samples should have means between what two numbers?
A)$2272.32 and $2727.68
B)$2,304.04 and $2,695.96
C)$150.00 and $2,695.96
D)$150.00 and $2,597.98
E)$300 and $4900

Accepted Answer

The mean of the population is $2,500 and the standard deviation is $250. The standard error of the mean for a sample of 150 students is $250 / sqrt(150) = $12.25. 99.7% of the samples should have means between $2,500 +/- 2.576 * $12.25 = $2,500 +/- $31.52, which is between $2272.32 and $2727.68.

Question 7

A hair stylist believes the distribution of her tips is Normal,with a mean of $10.50 and a standard deviation of $6.25.What is the probability that her 15 clients this weekend will tip an average of at least $12?
A)0.8238
B)0.2371
C)0.4052
D)0.5948
E)0.1762

Accepted Answer

To find the probability that the average tip from 15 clients is at least $12, we use the formula for the z-score: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the values, we get $z = \frac{12 - 10.5}{6.25/\sqrt{15}}$. This gives a z-score, which we then look up in a standard normal distribution table or use a calculator to find the corresponding probability. The correct answer reflects the probability of getting an average tip of at least $12, which is found by subtracting the cumulative probability up to the calculated z-score from 1.

Question 8

The weight of crackers in a box is stated to be 450 grams.The amount that the packaging machine puts in the boxes is believed to have a Normal model with mean 465 grams and standard deviation 30 grams.What is the probability that the mean weight of a 50-box case of crackers is above 450 grams?
A)0.0002
B)0.9996
C)0.9998
D)0.0004
E)0.9994

Accepted Answer

We need to calculate the probability that the mean weight of a 50-box case of crackers is above 450 grams, given that the packaging machine puts in the boxes with a mean of 465 grams and a standard deviation of 30 grams.
 
To do this, we can use the Central Limit Theorem, which tells us that the distribution of sample means will be approximately normal with a mean of 465 grams and a standard deviation of 30/sqrt(50) grams.

Z = (450 - 465) / (30/sqrt(50)) = -5sqrt(50) / 3
 
From the standard normal distribution table, the area to the right of -5.09 is 0.9998. Therefore, the probability that the mean weight of a 50-box case of crackers is above 450 grams is 0.9998. 
 
Hence, the answer is C) 0.9998.

Question 9

A candy company claims that 17% of the jelly beans in its spring mix are pink.Suppose that the candies are packaged at random in bags containing about 400 jelly beans.A class of students opens several bags,counts the various colours of jelly beans,and calculates the proportion that are pink.In one bag,the students found 14% of the jelly beans were pink.Is this an unusually small proportion of pink jelly beans? Explain your response.
A)This is an extremely unlikely result.It is 0.21 standard deviations below the mean.
B)This is a very unusual result.It is 0.08 standard deviations below the mean.
C)This is not an unusual result.It is only 1.60 standard deviations below the mean.
D)This is a very unusual result.It is 0.05 standard deviations below the mean.
E)This is not an unusual result.It is only 0.05 standard deviations below the mean.

Accepted Answer

This is not an unusual result because being 1.60 standard deviations below the mean is within the range of what might be expected due to natural sampling variability. In a normal distribution, results within 2 standard deviations from the mean are considered not unusual.

Question 10

In a large class,the professor has each person toss a coin 200 times and calculate the proportion of his or her tosses that were tails.The students then report their results,and the professor records the proportions.One student claims to have tossed her coin 200 times and found 58% tails.What do you think of this claim? Explain your response.
A)This is a fairly unusual result.Her proportion is about 1.60 standard deviations above the mean.
B)This is a typical result.Her proportion is only 2.26 standard deviations above the mean.
C)This is a fairly unusual result.Her proportion is about 2.26 standard deviations above the mean.
D)This is an extremely unlikely result.Her proportion is about 200 standard deviations above the mean.
E)This is a typical result.Her proportion is only 1.60 standard deviations above the mean.

Accepted Answer

The expected proportion of tails in a coin toss is 0.5. The standard deviation of the proportion of tails in 200 coin tosses is sqrt(0.5*0.5/200) ~= 0.035. Therefore, a proportion of 0.58 is about (0.58-0.5)/0.035 = 2.26 standard deviations above the mean. This is a fairly unusual result, but not extremely unlikely.

Question 11

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of less than $50 on dinner?
A)0.4121
B)0.9868
C)0.0132
D)0.5879
E)0.9614

Accepted Answer

According to the central limit theorem, the average of a large sample (in this case, 100 customers) will follow a normal distribution with a mean of $54 and a standard deviation of $1.8 ($18/sqrt(100)). We need to find the probability that this average is less than $50, which is equivalent to finding the z-score of $50 using the normal distribution with mean $54 and standard deviation $1.8.

z-score = (50-54)/1.8 = -2.22

Using a standard normal distribution table, we find that the probability of a z-score being less than -2.22 is 0.0132. Therefore, the probability that the next 100 customers will spend an average of less than $50 on dinner is 0.0132.

Question 12

Statistics from a weather centre indicate that a certain city receives an average of 23 inches of snow each year,with a standard deviation of 3 inches.Assume that a Normal model applies.A student lives in this city for 4 years.Let 
Represent the mean amount of snow for those 4 years.Describe the sampling distribution model of this sample mean.
A)Binom(23,3)
B)N(23,1.5)
C)N(23,0.75)
D)N(23,3)
E)There is not enough information to describe the distribution.

Accepted Answer

The answer of Statistics from a weather centre indicate that...

Question 13

Some real estate specialists estimate that the length of time people live in a house has a mean of 10 years and a standard deviation of 3 years.A random sample of 200 families was chosen and surveyed.Let 
Represent the mean number of years that those families had lived in their house.Describe the sampling distribution model of this mean.
A)Binom(10,3)
B)N(10,1.5)
C)N(10,3)
D)N(10,0.12)
E)There is not enough information to describe the distribution.

Accepted Answer

The answer of Some real estate specialists estimate that the...

Question 14

The number of hours per week that high school students spend on computers is normally distributed,with a mean of 5 hours and a standard deviation of 2 hours.70 students are chosen at random.Let 
Represent the mean number of hours spent on the computer for this group.Find the probability that 
Is between 5.1 and 5.7.
A)0.672
B)0.660
C)0.115
D)0.336
E)2.510

Accepted Answer

The answer of The number of hours per week that...

Question 15

Researchers believe that 7% of children have a gene that may be linked to a certain childhood disease.In an effort to track 50 of these children,researchers test 950 newborns for the presence of this gene.What is the probability that they do not find enough subjects for their study?
A)0.9821
B)0.0179
C)0.0358
D)0.2117
E)0.9581

Accepted Answer

The probability of selecting a child with the gene is 7/100=0.07. The probability of selecting a child without the gene is 1-0.07=0.93. The probability that none of the 950 newborns have the gene is (0.93)^950 which is approximately equal to 0.0179. Therefore, the probability that they do not find enough subjects for their study is 0.0179 or 1.79%. Thus, the answer is B.

Question 16

If we imagine all the possible random samples of 150 students at this university,95% of the samples should have means between what two numbers?
A)$2,206.06 and $2,793.94
B)$150.00 and $2,695.96
C)$2,304.04 and $2,695.96
D)$300 and $4900
E)$150.00 and $2,597.98

Accepted Answer

By the Central Limit Theorem, the distribution of sample means for samples of size 150 will be approximately normal with a mean equal to the population mean and a standard deviation of $\frac{\sigma}{\sqrt{150}}$, where $\sigma$ is the population standard deviation. Since we do not know the population standard deviation, we use the sample standard deviation, which is given as $500.00. Therefore the standard deviation of the sample means is $\frac{500.00}{\sqrt{150}} \approx 40.82$. 
To find the range of sample means that capture 95% of the possible values, we look at the 2.5th and 97.5th percentiles of a normal distribution with mean $2500.00$ and standard deviation $40.82$:

$$	ext{Lower bound: } 2500.00 - 1.96(40.82) \approx 2304.04$$
$$	ext{Upper bound: } 2500.00 + 1.96(40.82) \approx 2695.96$$

Thus, 95% of the possible random samples should have means between $\$2,304.04$ and $\$2,695.96$. Option C is the only one that contains these numbers.

Question 17

Automatic teller machine receipts show that customer withdrawals have a skewed distribution with a mean of $72 and a standard deviation of $15.Let 
Represent the mean withdrawal for the next 200 customers at the machine.Describe the sampling distribution model of this mean.
A)The distribution is skewed with a mean of $72 and a standard deviation of $15.
B)N(72,15)
C)N(5.1,15)
D)N(72,1.41)
E)There is not enough information to describe the distribution.

Accepted Answer

The answer of Automatic teller machine receipts show that customer...

Question 18

A national study reported that 76% of high school graduates pursue a college education immediately after graduation.A private high school advertises that 156 of their 196 graduates last year went on to college.Does this school have an unusually high proportion of students going to university?
A)This school can boast an unusually high proportion of students going to university.Their proportion is 0.94 standard deviations above the mean.
B)This school cannot boast an unusually high proportion of students going to university.Their proportion is only 0.71 standard deviations above the mean.
C)This school cannot boast an unusually high proportion of students going to university.Their proportion is only 1.18 standard deviations above the mean.
D)This school cannot boast an unusually high proportion of students going to university.Their proportion is only 0.94 standard deviations above the mean.
E)This school can boast an unusually high proportion of students going to university.Their proportion is 1.41 standard deviations above the mean.

Accepted Answer

First, we need to calculate the mean and standard deviation of the proportion of high school graduates who go on to college immediately.

Mean = 76% or 0.76 
Standard deviation = sqrt(0.76 * 0.24) = 0.427

Next, we can calculate the z-score for the private high school's proportion:

z = (0.796 - 0.76) / 0.427 = 0.845

Finally, we can use a z-table to determine how many standard deviations above or below the mean this z-score corresponds to.

A z-score of 0.845 corresponds to an area of 0.7995 in the z-table. This means that the proportion of students from this private high school is 1.18 standard deviations above the mean.

Therefore, choice C is correct.

Question 19

Information on a packet of seeds claims that the germination rate is 75%.Should you be surprised if 141 of the 150 seeds in the packet germinate?
A)You should not be surprised.A germination rate of 94.0% is only 3.22 standard deviations above the mean.
B)You should be very surprised.A germination rate of 94.0% is 5.37 standard deviations above the mean.
C)You should not be surprised.A germination rate of 94.0% is only 5.37 standard deviations above the mean.
D)You should be very surprised.A germination rate of 94.0% is 8.06 standard deviations above the mean.
E)You should be very surprised.A germination rate of 94.0% is 6.72 standard deviations above the mean.

Accepted Answer

The mean germination rate for the packet of seeds is 75%, with a standard deviation of (0.75 x 0.25)^(1/2) = 0.433. To find out how many standard deviations 141 out of 150 seeds germinating is from the mean, we can use the formula:
z = (x - μ) / σ
where x is the observed value, μ is the mean, and σ is the standard deviation. Plugging in the values, we get:
z = (141/150 - 0.75) / 0.433
z = 5.37
This means that the observed germination rate of 94% is 5.37 standard deviations above the mean. According to standard normal distribution, such an event has a probability of less than 0.00001%. Therefore, you should be very surprised if 141 out of 150 seeds in the packet germinate.

Question 20

The number of hours per week that high school students spend on homework is normally distributed,with a mean of 12 hours and a standard deviation of 3 hours.50 students are chosen at random.Let 
Represent the mean number of hours spent on homework for this group.Find the probability that 
Is between 10.8 and 13.0.
A)0.0078
B)0.286
C)0.9886
D)0.4711
E)0.0143

Accepted Answer

We know that the distribution of the mean number of hours spent on homework follows a normal distribution with mean μ = 12 and standard deviation σ/√n = 3/√50 ≈ 0.4243.

To find the probability that the mean number of hours is between 10.8 and 13.0, we need to standardize these values:

z1 = (10.8 - 12) / 0.4243 ≈ -3.3842
z2 = (13 - 12) / 0.4243 ≈ 2.3591

Using a standard normal distribution table or calculator, we find the probability that z is between -3.3842 and 2.3591 is approximately 0.9886.

Therefore, the answer is choice C (0.9886).

Quiz 17: Sampling Distribution Models

Suppose that the national mean weight for a three-year-old girl is 30 pounds with a standard deviation of 5 pounds.A pediatric office takes a random sample of 100 three-year-old girls,records their weights,and finds the mean.Describe the sampling distribution model of this mean.

The weight of crackers in a box is stated to be 450 grams.The amount that the packaging machine puts in the boxes is believed to have a Normal model with mean 465 grams and standard deviation 30 grams.What is the probability that the mean weight of a 50-box case of crackers is below 450 grams?

Researchers believe that 6% of children have a gene that may be linked to a certain childhood disease.In an effort to track 50 of these children,researchers test 950 newborns for the presence of this gene.What is the probability that they find enough subjects for their study?

If we imagine all the possible random samples of 250 students at this university,68% of the samples should have means between what two numbers?

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of at least $58 on dinner?

If we imagine all the possible random samples of 150 students at this university,99.7% of the samples should have means between what two numbers?

A hair stylist believes the distribution of her tips is Normal,with a mean of $10.50 and a standard deviation of $6.25.What is the probability that her 15 clients this weekend will tip an average of at least $12?

The weight of crackers in a box is stated to be 450 grams.The amount that the packaging machine puts in the boxes is believed to have a Normal model with mean 465 grams and standard deviation 30 grams.What is the probability that the mean weight of a 50-box case of crackers is above 450 grams?

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of less than $50 on dinner?

Researchers believe that 7% of children have a gene that may be linked to a certain childhood disease.In an effort to track 50 of these children,researchers test 950 newborns for the presence of this gene.What is the probability that they do not find enough subjects for their study?

If we imagine all the possible random samples of 150 students at this university,95% of the samples should have means between what two numbers?

Automatic teller machine receipts show that customer withdrawals have a skewed distribution with a mean of $72 and a standard deviation of $15.Let Represent the mean withdrawal for the next 200 customers at the machine.Describe the sampling distribution model of this mean.

A national study reported that 76% of high school graduates pursue a college education immediately after graduation.A private high school advertises that 156 of their 196 graduates last year went on to college.Does this school have an unusually high proportion of students going to university?

Information on a packet of seeds claims that the germination rate is 75%.Should you be surprised if 141 of the 150 seeds in the packet germinate?