Question 1

Find the indicated probability.
-The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week?
A) 0.2177
B) 0.2823
C) 0.1003
D) 0.7823

Accepted Answer

We need to find the probability of a teacher earning more than $525 a week. First we standardize the value using the formula z = (x - μ) / σ where x is the value ($525), μ is the mean ($490) and σ is the standard deviation ($45). 
z = (525 - 490) / 45 = 0.7778
Using a standard normal table or calculator, the probability of getting a value greater than 0.7778 is approximately 0.2177. So the answer is choice A.

Question 2

Find the indicated probability.
-The volumes of soda in quart soda bottles are normally distributed with a mean of 32.3 oz and a standard deviation of 1.2 oz. What is the probability that the volume of soda in a randomly selected bottle will be less
Than 32 oz?
A) 0.0987
B) 0.5987
C) 0.3821
D) 0.4013

Accepted Answer

The answer of Find the indicated probability.
-The volumes of soda...

Question 3

Solve the problem.
-The systolic blood pressures of the patients at a hospital are normally distributed with a mean of 138 mm Hg and a standard deviation of 13.5 mm Hg. Find the two blood pressures having these properties: the mean is
Midway between them and 90% of all blood pressures are between them.
A) 120.7 mm Hg, 155.3 mm Hg
B) 115.8 mm Hg, 160.2 mm Hg
C) 125.9 mm Hg, 150.2 mm Hg
D) 116.8 mm Hg, 161.2 mm Hg

Accepted Answer

We need to find the blood pressures that are midway between them and contain 90% of all blood pressures.

Using the empirical rule, we know that 90% of the data falls within 1.645 standard deviations from the mean.

To find the two blood pressures, we can use the formula:

Lower bound = mean - (z-score)(standard deviation)
Upper bound = mean + (z-score)(standard deviation)

Let x be the blood pressure we are looking for. We know that:

mean = (x1 + x2)/2, where x1 and x2 are the two blood pressures
x1 = 2(mean) - x2

Substituting the values we know:

mean = 138
standard deviation = 13.5
z-score = 1.645

Solving for x2:

138 + 1.645(13.5) = x2

x2 = 160.2 mm Hg

Using x1 = 2(mean) - x2:

x1 = 115.8 mm Hg

Therefore, the two blood pressures that are midway between them and contain 90% of all blood pressures are 115.8 mm Hg and 160.2 mm Hg, which correspond to choice B.

Question 4

Find the indicated probability.
-The diameters of pencils produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What is the probability that the diameter of a randomly selected pencil
Will be less than 0.285 inches?
A) 0.4332
B) 0.0596
C) 0.0668
D) 0.9332

Accepted Answer

The answer of Find the indicated probability.
-The diameters of pencils...

Question 5

Solve the problem.
-A poll of 1200 randomly selected students in grades 6 through 8 was conducted and found that 59% enjoy
playing sports. What is the sampling distribution suggested by the given data?

Accepted Answer

The answer of Solve the problem.
-A poll of 1200 randomly...

Question 6

Solve the problem.
-Scores on a test have a mean of 67 and $\mathrm { Q } _ { 3 }$ is 83. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.
A) 21.3
B) 12
C) 10.7
D) 23.9

Accepted Answer

The answer of Solve the problem.
-Scores on a test have...

Question 7

Solve the problem.
-A history teacher assigns letter grades on a test according to the following scheme: 
A: Top 10\%
B: Scores below the top $10 \%$ and above the bottom $60 \%$
C: Scores below the top 40\% and above the bottom 20\%
D: Scores below the top $80 \%$ and above the bottom $10 \%$
F: Bottom $10 \%$ Scores on the test are normally distributed with a mean of 70 and a standard deviation of 12.9. Find the numerical limits for each letter grade. 
A)A: Above 82
B: Between 73 and 82
C: Between 60 and 72
D: Between 53 and 59
F: Below 53

B)A: Above 82
B: Between 73 and 82
C: Between 60 and 72
D: Between 56 and 59
F: Below 56

C)A: Above 87
B: Between 77 and 87
C: Between 59 and 76
D: Between 53 and 58
F: Below 53

D) A: Above 87
B: Between 73 and 87
C: Between 59 and 72
D: Between 53 and 58
F: Below 53

Accepted Answer

The answer of Solve the problem.
-A history teacher assigns letter...

Question 8

Assume that X has a normal distribution, and find the indicated probability.
-The mean is $\mu = 137.0$ and the standard deviation is $\sigma = 5.3 .$ Find the probability that X is between 134.4 and 140.1.
A) 0.8138
B) 0.6242
C) 1.0311
D) 0.4069

Accepted Answer

The answer of Assume that X has a normal distribution,...

Question 9

Solve the problem.
-A recent survey based on a random sample of n = 450 voters, predicted that the Independent candidate for the
mayoral election will get 24% of the vote, but he actually gets 28%. Can it be concluded that the survey was
done incorrectly?

Accepted Answer

The answer of Solve the problem.
-A recent survey based on...

Question 10

Solve the problem.
-A math teacher gives two different tests to measure students' aptitude for math. Scores on the first test are normally distributed with a mean of 22 and a standard deviation of 5. Scores on the second test are normally
Distributed with a mean of 71 and a standard deviation of 10.3. Assume that the two tests use different scales to
Measure the same aptitude. If a student scores 29 on the first test, what would be his equivalent score on the
Second test? (That is, find the score that would put him in the same percentile.)
A) 86
B) 83
C) 78
D) 85

Accepted Answer

The answer of Solve the problem.
-A math teacher gives two...

Question 11

Find the indicated probability.
-The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. What is the probability that a pregnancy lasts at least 300 days?
A) 0.4834
B) 0.0179
C) 0.9834
D) 0.0166

Accepted Answer

To solve the problem, we use the z-score formula:

z = (x - mu) / sigma

where x is the value we're interested in (in this case, 300), mu is the mean (268), and sigma is the standard deviation (15).

Plugging in the numbers, we get:

z = (300 - 268) / 15 = 2.13333

Now we look up the probability corresponding to a z-score of 2.13333 in a standard normal distribution table. The probability is about 0.0166.

Therefore, the probability that a pregnancy lasts at least 300 days is 0.0166, or approximately 1.66%. This corresponds to answer choice D.

Question 12

Solve the problem.
-A poll of 2000 randomly selected students in grades 6 through 8 was conducted and found that 41% enjoy
playing sports. Would confidence in the results increase if the sample size were 3800 instead of 2000? Why or
why not?

Accepted Answer

The answer of Solve the problem.
-A poll of 2000 randomly...

Question 13

Find the indicated probability.
-A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200
And 275.
A) 0.4332
B) 0.5
C) 0.0668
D) 0.9332

Accepted Answer

To find the probability of a rating that is between 200 and 275, we need to find the area under the normal curve between the z-scores that correspond to these values.

The z-score for a rating of 200 is:

z = (200 - 200) / 50 = 0

And the z-score for a rating of 275 is:

z = (275 - 200) / 50 = 1.5

Using a standard normal distribution table or calculator, we can find that the area under the curve between z = 0 and z = 1.5 is approximately 0.4332. Therefore, the probability of a rating between 200 and 275 is 0.4332.

Question 14

Find the indicated probability.
-A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 170
And 220.
A) 0.1554
B) 0.0703
C) 0.3811
D) 0.2257

Accepted Answer

We need to find the probability of getting a rating between 170 and 220, which is equivalent to finding the area under the normal distribution curve between 170 and 220. We can standardize the values using the formula z = (x - μ) / σ, where x is the rating, μ is the mean, and σ is the standard deviation.

For 170, z = (170 - 200) / 50 = -0.6
For 220, z = (220 - 200) / 50 = 0.4

Using a standard normal distribution table or calculator, we can find the probabilities associated with these z-scores:
P(z < -0.6) = 0.2743
P(z < 0.4) = 0.6554

The probability of getting a rating between 170 and 220 is the difference between these two probabilities:
P(170 < x < 220) = P(-0.6 < z < 0.4) = 0.6554 - 0.2743 = 0.3811

Therefore, the answer is C, 0.3811.

Question 15

Find the indicated probability.
-Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal
Quarters will be rejected?
A) 0.0196%
B) 2.48%
C) 1.96%
D) 1.62%

Accepted Answer

To find the percentage of legal quarters that will be rejected, we need to find the proportion of quarters that weigh outside the range of 5.48 g and 5.82 g. We can convert these limits into z-scores using the formula z = (x - μ) / σ, where x is the weight of the coin, μ is the mean weight, and σ is the standard deviation. For the lower limit, we have z = (5.48 - 5.67) / 0.070 = -2.71. For the upper limit, we have z = (5.82 - 5.67) / 0.070 = 2.14. We can use a standard normal table or calculator to find the area to the left of -2.71 and the area to the right of 2.14, and add them together to get the proportion of quarters that will be rejected. This comes out to be approximately 1.96%, which corresponds to choice C.

Question 16

Find the indicated probability.
-The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What percentage of bolts will have a diameter greater than 0.32 inches?
A) 47.72%
B) 97.72%
C) 37.45%
D) 2.28%

Accepted Answer

We need to find the probability that a bolt has a diameter greater than 0.32 inches.

Using the z-score formula: z = (x - μ) / σ

z = (0.32 - 0.30) / 0.01 = 2

Lookup in the z-table or use a calculator, we find that the area to the right of z = 2 is 0.0228 or 2.28%.

Therefore, the percentage of bolts that will have a diameter greater than 0.32 inches is 2.28%.

The correct choice is D.

Question 17

Find the indicated probability.
-In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. For a randomly selected home, find
The probability that the September energy consumption level is between 1100 kWh and 1225 kWh.
A) 0.0910
B) 0.2881
C) 0.1971
D) 0.3791

Accepted Answer

Using the normal distribution formula, we can calculate the z-scores for the given values:
z1 = (1100 - 1050) / 218 = 0.229
z2 = (1225 - 1050) / 218 = 0.804
Then, we can find the area under the standard normal distribution curve between these two z-scores using a standard normal distribution table or a calculator:
P(0.229 < z < 0.804) = 0.1971
Therefore, the probability that the September energy consumption level is between 1100 kWh and 1225 kWh is 0.1971, which corresponds to option C.

Question 18

Find the indicated probability.
-The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. What percentage of trainees earn less than $900 a month?
A) 40.82%
B) 35.31%
C) 9.18%
D) 90.82%

Accepted Answer

To find the probability that a trainee earns less than $900 a month, we need to find the z-score for $900 using the formula:

z = (x - μ) / σ = (900 - 1100) / 150 = -1.33

Using a standard normal distribution table, we can find that the probability of a z-score less than -1.33 is 0.0918, or 9.18%. Therefore, choice C is the best answer.

Question 19

Solve the problem.
-A poll of 1200 randomly selected students in grades 6 through 8 was conducted and found that 34% enjoy
playing sports. Is the 34% result a statistic or a parameter? Explain.

Accepted Answer

The answer of Solve the problem.
-A poll of 1200 randomly...

Question 20

Solve the problem.
-Scores on a test have a mean of 70 and $\mathrm { Q } _ { 3 }$ is 83. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.)
A) 93.7
B) 95.6
C) 94.8
D) 92.9

Accepted Answer

The z-score for Q3 is (83-70)/$\sigma$=1.3. Using a standard normal distribution table, we find that the z-score for P90 is 1.28. Therefore, P90 = 70 + 1.28$\sigma$ = 94.8.

Quiz 6: Normal Probability Distributions

Find the indicated probability. -The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a standard deviation of $45. What is the probability that a randomly selected teacher earns more than $525 a week?

Find the indicated probability. -The volumes of soda in quart soda bottles are normally distributed with a mean of 32.3 oz and a standard deviation of 1.2 oz. What is the probability that the volume of soda in a randomly selected bottle will be less Than 32 oz?

Find the indicated probability. -The diameters of pencils produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What is the probability that the diameter of a randomly selected pencil Will be less than 0.285 inches?

Solve the problem. -A poll of 1200 randomly selected students in grades 6 through 8 was conducted and found that 59% enjoy playing sports. What is the sampling distribution suggested by the given data?

Solve the problem. -Scores on a test have a mean of 67 and Q3\mathrm { Q } _ { 3 }Q3​ is 83. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.

Assume that X has a normal distribution, and find the indicated probability. -The mean is μ=137.0\mu = 137.0μ=137.0 and the standard deviation is σ=5.3.\sigma = 5.3 .σ=5.3. Find the probability that X is between 134.4 and 140.1.

Solve the problem. -A recent survey based on a random sample of n = 450 voters, predicted that the Independent candidate for the mayoral election will get 24% of the vote, but he actually gets 28%. Can it be concluded that the survey was done incorrectly?

Find the indicated probability. -The lengths of human pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. What is the probability that a pregnancy lasts at least 300 days?

Solve the problem. -A poll of 2000 randomly selected students in grades 6 through 8 was conducted and found that 41% enjoy playing sports. Would confidence in the results increase if the sample size were 3800 instead of 2000? Why or why not?

Find the indicated probability. -A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200 And 275.

Find the indicated probability. -A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 170 And 220.

Find the indicated probability. -Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal Quarters will be rejected?

Find the indicated probability. -The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviation of 0.01 inches. What percentage of bolts will have a diameter greater than 0.32 inches?

Find the indicated probability. -The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of $150. What percentage of trainees earn less than $900 a month?

Solve the problem. -A poll of 1200 randomly selected students in grades 6 through 8 was conducted and found that 34% enjoy playing sports. Is the 34% result a statistic or a parameter? Explain.

Solve the problem. -Scores on a test have a mean of 70 and Q3\mathrm { Q } _ { 3 }Q3​ is 83. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.)

Solve the problem. -Scores on a test have a mean of 67 and $\mathrm { Q } _ { 3 }$ is 83. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.

Assume that X has a normal distribution, and find the indicated probability. -The mean is $\mu = 137.0$ and the standard deviation is $\sigma = 5.3 .$ Find the probability that X is between 134.4 and 140.1.

Solve the problem. -Scores on a test have a mean of 70 and $\mathrm { Q } _ { 3 }$ is 83. The scores have a distribution that is approximately normal. Find P90. (You will need to first find the standard deviation.)