Question 1

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours.What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours?
A)0.2412
B)0.3830
C)0.1712
D)0.5121

Accepted Answer

To solve this problem, we need to find the area under the normal curve between 70 and 80 hours. We can do this by calculating the z-scores for each value, using the formula:

z = (x - mu) / sigma

where x is the value (in this case, either 70 or 80), mu is the mean (75), and sigma is the standard deviation (10). Plugging in the numbers, we get:

z(70) = (70 - 75) / 10 = -0.5
z(80) = (80 - 75) / 10 = 0.5

Now, we can use a standard normal table (or a calculator) to find the area between these two z-scores. The area between -0.5 and 0.5 is 0.3830, so the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours is 0.3830. Therefore, the best choice is B.

Question 2

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P(  ≥ 0.27).
A)0.7422
B)0.8141
C)0.6125
D)0.6841

Accepted Answer

The answer of A population has a proportion equal to...

Question 3

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours.If the manufacturer of the battery is able to reduce the standard deviation of battery life from 10 to 9 hours,what would be the probability that 16 batteries randomly sampled from the population will have a sample mean life of between 70 and 80 hours?
A)0.6127
B)0.8124
C)0.9736
D)0.8812

Accepted Answer

We can use the central limit theorem and the standard normal distribution to solve this problem.
First, we need to standardize the sample mean, which follows a normal distribution with mean = 75 and standard deviation = 9/sqrt(16) = 2.25. Thus, we have:
z = (x̄ - μ) / (σ / sqrt(n)) = (70 - 75) / (2.25) = -2.22
z = (x̄ - μ) / (σ / sqrt(n)) = (80 - 75) / (2.25) = 2.22
Next, we use a standard normal distribution table (or calculator) to find the probabilities associated with these z-scores:
P(z < -2.22) = 0.0139
P(z < 2.22) = 0.9861
Finally, we subtract these probabilities to find the probability that the sample mean falls between 70 and 80 hours:
P(-2.22 < z < 2.22) = P(z < 2.22) - P(z < -2.22) = 0.9861 - 0.0139 = 0.9736

Therefore, the answer is C (0.9736).

Question 4

In an article entitled "Fuel Economy Calculations to Be Altered," James R.Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks.This is the first modification since 1985.It is expected to lower average mileage for city driving in conventional cars from 10% to 20%.AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg).The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg.Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg.Suppose 100 AAA members were to test the 2008 F-150.Determine the probability that the average mpg would be at least 15.7.
A)0.0026
B)0.0121
C)0.0451
D)0.0001

Accepted Answer

The answer of In an article entitled "Fuel Economy Calculations...

Question 5

A random sample of 100 items is selected from a population of size 350.What is the probability that the sample mean will exceed 200 if the population mean is 195 and the population standard deviation equals 20? (Hint: Use the finite correction factor since the sample size is more than 5% of the population size. )
A)0.0415
B)0.0016
C)0.0241
D)0.0171

Accepted Answer

The answer of A random sample of 100 items is...

Question 6

In an article entitled "Fuel Economy Calculations to Be Altered," James R.Healey indicated that the government planned to change how it calculates fuel economy for new cars and trucks.This is the first modification since 1985.It is expected to lower average mileage for city driving in conventional cars from 10% to 20%.AAA has forecast that the 2008 Ford F-150 would achieve 15.7 mile per gallon (mpg).The 2008 Ford F-150 was tested by AAA members driving the vehicle themselves and was found to have an average of 14.3 mpg.Assume that the mean obtained by AAA members is the true mean for the population of 2008 Ford F-150 trucks and that the population standard deviation is 5 mpg.The current method of calculating the mpg forecasts that the 2008 F-150 will average 16.8 mpg.Determine the probability that these same 100 AAA members would average more than 16.8 mpg while testing the 2008 F-150.
A)0
B)0.0155
C)0.0412
D)None of the above

Accepted Answer

Since the true mean for the population is found to be 14.3 mpg and the population standard deviation is given as 5 mpg, we can calculate the standard error of the mean as follows:
Standard Error of the Mean = (Population Standard Deviation) / sqrt(sample size)
= 5 / sqrt(100)
= 0.5 mpg
Now, we can calculate the z-score for this sample mean as follows:
z = (sample mean - population mean) / standard error of the mean
= (16.8 - 14.3) / 0.5
= 5
This z-score is very high and indicates that the sample mean is very unlikely to occur if the population mean is 16.8 mpg. Therefore, the probability of getting a sample mean greater than 16.8 mpg is practically zero, so the answer is A.

Question 7

Suppose that a population is known to be normally distributed with mean = 2,000 and standard deviation = 230.If a random sample of size n = 8 is selected,calculate the probability that the sample mean will exceed 2,100.
A)0.2141
B)0.1871
C)0.0712
D)0.1093

Accepted Answer

The answer of Suppose that a population is known to...

Question 8

A normally distributed population has a mean of 500 and a standard deviation of 60.Determine the probability that a random sample of size 16 selected from this population will have a sample mean less than 475.
A)0.3251
B)0.7124
C)0.0475
D)0.0712

Accepted Answer

To solve this problem, we use the formula for the distribution of sample means: 
μx̄ = μ = 500 and σx̄ = σ/√n = 60/√16 = 15
Now, we need to find the probability that a sample mean is less than 475, which can be written as: 
P(x̄ < 475) = P((x̄ - μ) / (σx̄) < (475 - 500) / 15) 
P(z < -1.67)
Using a standard normal table or calculator, we find that the probability of getting a z-score less than -1.67 is 0.0475. Therefore, the answer is C.

Question 9

Given a population where the proportion of items with a desired attribute is p = 0.25,if a sample of 400 is taken,what is the standard deviation of the sampling distribution of  ?
A)0.0512
B)0.0312
C)0.0217
D)0.0412

Accepted Answer

The standard deviation of the sampling distribution of a proportion is given by the formula:
SD = sqrt [(p(1-p))/n], where p is the population proportion of items with the desired attribute, and n is the sample size.

Substituting the given values, we get:
SD = sqrt [(0.25 x 0.75)/400]
SD = sqrt (0.00046875)
SD = 0.0217 (rounded to 4 decimal places)

Therefore, the answer is C.

Question 10

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P(  > 0.40).
A)0.0146
B)0.0411
C)0.0521
D)0.0312

Accepted Answer

The answer of A population has a proportion equal to...

Question 11

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours.What is the probability that 16 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours?
A)0.9444
B)0.5121
C)0.7124
D)0.1512

Accepted Answer

To find the probability that the sample mean life of 16 batteries is between 70 and 80 hours, we use the formula for the standard error of the mean: $SE = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 10$ and $n = 16$, so $SE = \frac{10}{\sqrt{16}} = \frac{10}{4} = 2.5$. Next, we calculate the Z-scores for 70 and 80 hours: $Z_{70} = \frac{70 - 75}{2.5} = -2$ and $Z_{80} = \frac{80 - 75}{2.5} = 2$. Looking up these Z-scores in a standard normal distribution table or using a calculator, we find the probabilities corresponding to Z = -2 and Z = 2, and the area between these Z-scores represents the probability we're interested in, which is approximately 0.9544. However, due to rounding or a typo in the options, the closest provided option is A) 0.9444.

Question 12

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P(  ≤ 0.35).
A)0.7244
B)0.8621
C)0.7124
D)0.6126

Accepted Answer

We can use the normal approximation to the binomial distribution:
mean = np = 100 x 0.3 = 30
standard deviation = sqrt(npq) = sqrt(100 x 0.3 x 0.7) = 4.58
z-score = (0.35 - 0.3) / 0.458 = 1.09
Using a standard normal table or a calculator, we find the probability to be approximately 0.8621. Therefore, the best choice is (B).

Question 13

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P(0.25 <  ≤ 0.40).
A)0.8121
B)0.7415
C)0.8475
D)0.5612

Accepted Answer

The answer of A population has a proportion equal to...

Question 14

A normally distributed population has a mean of 500 and a standard deviation of 60.Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515.
A)0.1056
B)0.1761
C)0.0712
D)0.0151

Accepted Answer

The answer of A normally distributed population has a mean...

Question 15

Given a population where the proportion of items with a desired attribute is p = 0.25,if a sample of 400 is taken,what is the probability the proportion of successes in the sample will be greater than 0.22?
A)0.9162
B)0.8812
C)0.7141
D)0.8412

Accepted Answer

This problem can be solved using the normal approximation to the binomial distribution. The mean of the sampling distribution of the proportion is $p = 0.25$, and the standard deviation is $\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.25(0.75)}{400}} = 0.02165$. We want to find the probability that the sample proportion is greater than 0.22, so we calculate the z-score for 0.22: $z = \frac{0.22 - 0.25}{0.02165} = -1.387$. Looking up this z-score in a standard normal distribution table or using a calculator gives us the probability of being less than 0.22. Since we want the probability of being greater, we subtract this value from 1. The exact probability might vary slightly depending on the method of calculation, but the closest answer provided is 0.9162, indicating a high likelihood of the sample proportion being greater than 0.22.

Question 16

The branch manager for United Savings and Loan in Seaside,Virginia,has worked with her employees in an effort to reduce the waiting time for customers at the bank.Recently,she and the team concluded that average waiting time is now down to 3.5 minutes with a standard deviation equal to 1.0 minute.However,before making a statement at a managers' meeting,this branch manager wanted to double-check that the process was working as thought.To make this check,she randomly sampled 25 customers and recorded the time they had to wait.She discovered that mean wait time for this sample of customers was 4.2 minutes.Based on the team's claims about waiting time,what is the probability that a sample mean for n = 25 people would be as large or larger than 4.2 minutes?
A)0.0214
B)0.0512
C)0.0231
D)0.0011

Accepted Answer

The answer of The branch manager for United Savings and...

Question 17

Suppose nine items are randomly sampled from a normally distributed population with a mean of 100 and a standard deviation of 20.The nine randomly sampled values are:  Calculate the probability of getting a sample mean that is smaller than the sample mean for these nine sampled values.
A)0.1411
B)0.1612
C)0.1512
D)0.2266

Accepted Answer

The answer of Suppose nine items are randomly sampled from...

Question 18

If a random sample of 200 items is taken from a population in which the proportion of items having a desired attribute is p = 0.30,what is the probability that the proportion of successes in the sample will be less than or equal to 0.27?
A)0.0841
B)0.1011
C)0.1912
D)0.1762

Accepted Answer

The answer of If a random sample of 200 items...

Question 19

The proportion of items in a population that possess a specific attribute is known to be 0.70.If a simple random sample of size n = 100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.65,what is the sampling error?
A)-0.03
B)-0.05
C)0.08
D)0.01

Accepted Answer

Sampling error is the difference between the sample statistic (proportion of items in the sample that contain the attribute of interest) and the population parameter (proportion of items in the population that possess the specific attribute). Therefore, the sampling error can be calculated as:
sampling error = sample proportion - population proportion
sampling error = 0.65 - 0.70
sampling error = -0.05
The negative sign indicates that the sample proportion is lower than the population proportion. The closest choice to this value is B) -0.05, which is the correct answer.

Question 20

SeeClear Windows makes windows for use in homes and commercial buildings.The standards for glass thickness call for the glass to average 0.375 inches with a standard deviation equal to 0.050 inch.Suppose a random sample of n = 50 windows yields a sample mean of 0.392 inches.What is the probability if the windows meet the standards?
A)0.0612
B)0.0082
C)0.0015
D)0.0009

Accepted Answer

The answer of SeeClear Windows makes windows for use in...

Quiz 7: Introduction to Sampling Distributions

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours.What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours?

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P( ≥ 0.27) .

Suppose that a population is known to be normally distributed with mean = 2,000 and standard deviation = 230.If a random sample of size n = 8 is selected,calculate the probability that the sample mean will exceed 2,100.

A normally distributed population has a mean of 500 and a standard deviation of 60.Determine the probability that a random sample of size 16 selected from this population will have a sample mean less than 475.

Given a population where the proportion of items with a desired attribute is p = 0.25,if a sample of 400 is taken,what is the standard deviation of the sampling distribution of ?

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P( > 0.40) .

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours.What is the probability that 16 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours?

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P( ≤ 0.35) .

A population has a proportion equal to 0.30.Calculate the following probabilities with n = 100.Find P(0.25 < ≤ 0.40) .

A normally distributed population has a mean of 500 and a standard deviation of 60.Determine the probability that a random sample of size 25 selected from the population will have a sample mean greater than or equal to 515.

Given a population where the proportion of items with a desired attribute is p = 0.25,if a sample of 400 is taken,what is the probability the proportion of successes in the sample will be greater than 0.22?

Suppose nine items are randomly sampled from a normally distributed population with a mean of 100 and a standard deviation of 20.The nine randomly sampled values are: Calculate the probability of getting a sample mean that is smaller than the sample mean for these nine sampled values.

If a random sample of 200 items is taken from a population in which the proportion of items having a desired attribute is p = 0.30,what is the probability that the proportion of successes in the sample will be less than or equal to 0.27?

The proportion of items in a population that possess a specific attribute is known to be 0.70.If a simple random sample of size n = 100 is selected and the proportion of items in the sample that contain the attribute of interest is 0.65,what is the sampling error?