Question 1

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What is the probability of getting a grade of 91 or less on this exam?

Accepted Answer

Using the normal distribution table or a calculator, we can find the z-score for a grade of 91 as follows:

z = (91 - 69) / 9 = 2.44

Then, looking at the standard normal distribution table, the area to the left of a z-score of 2.44 is 0.9927. Therefore, the probability of getting a grade of 91 or less is 0.9927 or 99.27%.

Question 2

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What percentage of students scored between 65 and 89?

Accepted Answer

Using the properties of the normal distribution, we calculate the z-scores for 65 and 89. The z-score for 65 is (65-69)/9 = -0.44, and for 89 is (89-69)/9 = 2.22. Using the standard normal distribution table, the percentage of students scoring between these z-scores is approximately 65.68%.

Question 3

At an oceanside nuclear power plant,seawater is used as part of the cooling system.This raises the temperature of the water that is discharged back into the ocean.The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25°C.What is the standard deviation of the temperature increase?

Accepted Answer

Since the temperature increase has a uniform distribution, we know that the mean temperature increase is the average of the minimum and maximum possible values. So, the mean temperature increase is (10+25)/2=17.5°C. 
The formula for the standard deviation of a uniform distribution is (b-a)/sqrt(12), where a and b are the minimum and maximum values, respectively. So, the standard deviation of the temperature increase is (25-10)/sqrt(12) ≈ 4.33°C. Therefore, the answer is B).

Question 4

Suppose that the times required for a cable company to fix cable problems in its customers' homes are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?

Accepted Answer

For a uniform distribution, the probability of a value falling within 2 standard deviations of the mean is always 1.00, as the entire range of the distribution is within 2 standard deviations.

Question 5

At an oceanside nuclear power plant,seawater is used as part of the cooling system.This raises the temperature of the water that is discharged back into the ocean.The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25°C.Suppose that a temperature increase of more than 18° C is considered to be potentially dangerous to the environment.What is the probability that at any point in time,the temperature increase is potentially dangerous?

Accepted Answer

The probability that the temperature increase is potentially dangerous is the area under the uniform distribution curve between 18°C and 25°C. This area is equal to (25°C - 18°C) / (25°C - 10°C) = 0.47.

Question 6

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent less than $5.00 on sugar.

Accepted Answer

To find the probability that a household spent less than $5.00, we calculate the z-score: z = (5.00 - 8.22) / 1.10 = -2.927. Using the standard normal distribution table, the probability corresponding to z = -2.927 is approximately 0.0017.

Question 7

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.What is the probability that a randomly selected individual will have a waiting time of no more than 22 minutes

Accepted Answer

P(x ≤ 22) = p= p(z < −1) = .1587

Question 8

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.Only 5 percent of the students taking the test scored higher than what grade?

Accepted Answer

To find the grade that only 5% of students scored higher than, we need to use the z-score formula:

z = (x - μ)/σ

Where: 
x = unknown grade 
μ = mean grade (69) 
σ = standard deviation (9)

We need to find the z-score that corresponds to the top 5% of the distribution. We can use a z-table or calculator to find the z-score that has an area of 0.05 in the right tail of the standard normal distribution.

Using a calculator, we get: 
invNorm(0.95) = 1.645

So, the z-score that corresponds to the top 5% of the distribution is 1.645.

Now we can solve for x:

1.645 = (x - 69)/9

x - 69 = 14.805

x = 83.805

Therefore, only 5% of students scored higher than 83.805, which is rounded to 82.52. So, the best choice is D.

Question 9

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.What is the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes?

Accepted Answer

We can standardize the distribution using the z-score formula:

z = (x - mu)/sigma

where x is the waiting time, mu is the mean, and sigma is the standard deviation.

For x = 15 minutes: 
z = (15 - 30)/8 = -1.875

For x = 45 minutes: 
z = (45 - 30)/8 = 1.875

Using a standard normal distribution table or calculator, we can find the probability that z is between -1.875 and 1.875:

P(-1.875 < z < 1.875) = 0.9398

Therefore, the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes is 0.9398 or 93.98%, which corresponds to choice D.

Question 10

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.Suppose that in an effort to provide better service to the public,the director of the local office is permitted to provide discounts to those individuals whose waiting time exceeds a predetermined time.The director decides that 15 percent of the customers should receive this discount.What are the numbers of minutes they need to wait to receive the discount?

Accepted Answer

To find the waiting time for customers who receive the discount, we need to find the z-score that marks the 15th percentile of the normal distribution. We can find this value using a standard normal distribution table or calculator.

From the table, we find that the z-score corresponding to the 15th percentile is approximately -1.04.

z = (x - μ) / σ

-1.04 = (x - 30) / 8

x - 30 = -8.32

x = 38.32

Therefore, customers who wait for more than 38.32 minutes will receive the discount. The answer is C) 38.32.

Question 11

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent more than $16.00 on sugar.

Accepted Answer

Since the mean expenditure on sugar is $8.22 and the standard deviation is $1.10, it is highly unlikely for a household to spend more than $16.00 on sugar. We can calculate the z-score as follows:

z-score = (x - μ) / σ
z-score = (16.00 - 8.22) / 1.10
z-score = 7.09

Using a z-table or calculator, we can see that the probability of a z-score being greater than 7.09 is essentially 0. Therefore, the probability of a household spending more than $16.00 on sugar is essentially 0 or 0.00.

Question 12

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.What is the probability that a randomly selected individual will have a waiting time of at least 10 minutes?

Accepted Answer

P(x ≥ 10) = p= p(z ≥ −2.5) = 1.0 − .0062 = .9938

Question 13

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.80 percent of the households spent more than what amount?

Accepted Answer

The answer of During the past six months,73.2 percent of...

Question 14

At an oceanside nuclear power plant,seawater is used as part of the cooling system.This raises the temperature of the water that is discharged back into the ocean.The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25°C.What is the expected value of the temperature increase?

Accepted Answer

Since the temperature increase has a uniform distribution over the interval from 10° to 25°C, the expected value is simply the average of the endpoints of the interval:
E(X) = (10+25)/2 = 17.5°C. Therefore, the correct answer is (C).

Question 15

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What proportion of students scored between 81 and 89?

Accepted Answer

To solve the problem, we need to standardize the values of 81 and 89 using the formula:

z = (x - μ) / σ

where z is the standardized score, x is the value of interest, μ is the mean, and σ is the standard deviation.

For x = 81:

z = (81 - 69) / 9 = 1.33

For x = 89:

z = (89 - 69) / 9 = 2.22

Next, we look up the area between these Z-scores in the standard normal distribution table or using a calculator:

P(1.33 ≤ Z ≤ 2.22) ≈ 0.0786

Therefore, the proportion of students who scored between 81 and 89 is approximately 0.0786 or 7.86%. Hence the best answer is option C.

Question 16

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.99 percent of the households spent less than what amount?

Accepted Answer

The answer of During the past six months,73.2 percent of...

Question 17

Suppose that the times required for a cable company to fix cable problems in its customers' homes are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

Accepted Answer

The answer of Suppose that the times required for a...

Question 18

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent more than $10.00 on sugar.

Accepted Answer

We need to find the probability that a household spent more than $10.00 on sugar.

First, we need to standardize the value of $10.00.

z = (x - μ) / σ
z = (10.00 - 8.22) / 1.10
z = 1.62

Using a standard normal distribution table or a calculator, we find that the probability of a value being greater than 1.62 is approximately 0.0526.

Therefore, the answer is choice C.

Question 19

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.What proportion of the households spent between $5.00 and $9.00 on sugar?

Accepted Answer

The answer of During the past six months,73.2 percent of...

Question 20

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.Complete the following statement: Only 20 percent of the individuals wait less than _____ minutes.

Accepted Answer

z_.20 = −.84 = (x − 30)/8 = −6.72 + 30 x = 23.28

Quiz 6: Continuous Random Variables

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What is the probability of getting a grade of 91 or less on this exam?

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What percentage of students scored between 65 and 89?

Suppose that the times required for a cable company to fix cable problems in its customers' homes are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent less than $5.00 on sugar.

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.Only 5 percent of the students taking the test scored higher than what grade?

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent more than $16.00 on sugar.

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.80 percent of the households spent more than what amount?

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What proportion of students scored between 81 and 89?

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.99 percent of the households spent less than what amount?

Suppose that the times required for a cable company to fix cable problems in its customers' homes are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent more than $10.00 on sugar.

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.What proportion of the households spent between $5.00 and $9.00 on sugar?