Question 1

Provide an appropriate response. Assume that the variable under consideration has a density curve.
-Given that 40.5% of all possible observations of the variable are less than 12, determine the area under the density curve that lies to the right of 12. 
A)0.595
B)0.095
C)0.345
D)0.405

Accepted Answer

Since 40.5% of observations are less than 12, the area to the left of 12 under the density curve is 0.405. Therefore, the area to the right is 1 - 0.405 = 0.595.

Question 2

Use the empirical rule to solve the problem.
-The systolic blood pressure of 18-year-old women is normally distributed with a mean of 120 mmHg and a standard deviation of 12 mmHg. What percentage of 18-year-old women have a
Systolic blood pressure between 96 mmHg and 144 mmHg? 
A)99.99%
B)68.26%
C)95.44%
D)99.74%

Accepted Answer

The empirical rule states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations. In this case, the mean is 120 mmHg and the standard deviation is 12 mmHg. So, 95% of the data falls between 120 - 2(12) = 96 mmHg and 120 + 2(12) = 144 mmHg.

Question 3

Use a table of areas to find the specified area under the standard normal curve.
-The area that lies between -1.10 and -0.36 
A)0.2239
B)0.4951
C)-0.2237
D)0.2237

Accepted Answer

The area between -1.10 and -0.36 is the area to the left of -0.36 minus the area to the left of -1.10. Using a table of areas, we find that the area to the left of -0.36 is 0.3557 and the area to the left of -1.10 is 0.1320. Therefore, the area between -1.10 and -0.36 is 0.3557 - 0.1320 = 0.2237.

Question 4

Find the indicated probability or percentage for the normally distributed variable.
-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)1.62%
B)2.48%
C)0.0196%
D)1.96%

Accepted Answer

To find the percentage of quarters rejected, calculate the probability that a quarter's weight is less than 5.48 g or more than 5.82 g. Using the standard normal distribution, find the z-scores for 5.48 g and 5.82 g, then find the cumulative probabilities. Subtract the cumulative probability between these z-scores from 1 to get the rejection percentage, which is approximately 1.96%.

Question 5

Use a table of areas for the standard normal curve to find the required z-score.
-Find the z-score for which the area under the standard normal curve to its left is 0.96 
A)1.75
B)1.03
C)1.82
D)-1.38

Accepted Answer

The z-score corresponding to an area of 0.96 to the left under the standard normal curve is approximately 1.75.

Question 6

Use a table of areas to obtain the shaded area under the standard normal curve.
-  
A)$0.7198$
B) $0.2802$
C) $0.8599$
D) $0.1401$

Accepted Answer

The answer of Use a table of areas to obtain...

Question 7

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution.
-Estimate the probability of getting exactly 43 boys in 90 births. 
A)0.0159
B)0.0729
C)0.1628
D)0.0764

Accepted Answer

The normal approximation to the binomial distribution can be used here with the mean μ = np and the standard deviation σ = sqrt(np(1-p)), where n is the number of trials (births, in this case, 90) and p is the probability of a success (having a boy, assumed to be 0.5 for simplicity). Thus, μ = 90*0.5 = 45 and σ = sqrt(90*0.5*0.5) = sqrt(22.5) ≈ 4.7434. To find the probability of getting exactly 43 boys, we use the continuity correction and look for the probability of getting a result between 42.5 and 43.5. Using the Z-score formula Z = (X - μ) / σ, we find the Z-scores for 42.5 and 43.5, and then use the standard normal distribution to find the probabilities corresponding to these Z-scores. The difference between these probabilities gives an approximation of the probability of getting exactly 43 boys. The closest value to the calculated probability using this method is option D) 0.0764.

Question 8

Find the specified percentile, quartile, or decile.
-Scores on an English test are normally distributed with a mean of 32.7 and a standard deviation of 8.2. Find the 41st percentile. 
A)37.5
B)27.9
C)30.8
D)34.6

Accepted Answer

To find the 41st percentile of a normally distributed set of scores, you can use the Z-score formula or a statistical table or calculator designed for the normal distribution. The 41st percentile falls below the mean, indicating that it is less than the average score. Using the Z-score formula or a normal distribution calculator with the given mean (32.7) and standard deviation (8.2), you can find that the score corresponding to the 41st percentile is approximately 30.8.

Question 9

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution.
-Two percent of hair dryers produced at a certain plant are defective. Estimate the probability that of 10,000 randomly selected hair dryers, the number of defectives is between 195 and 210 inclusive. 
A)0.4251
B)0.4034
C)0.4017
D)0.3989

Accepted Answer

The normal distribution is a good approximation to the binomial distribution when np and n(1-p) are both at least 5. In this case, np = 10,000 * 0.02 = 200 and n(1-p) = 10,000 * 0.98 = 9800, so the normal distribution is a good approximation.The mean of the binomial distribution is np = 200 and the standard deviation is sqrt(np(1-p)) = sqrt(200 * 0.98) = 14.0.The probability that the number of defectives is between 195 and 210 inclusive is the area under the normal curve between z = (195 - 200) / 14 = -0.36 and z = (210 - 200) / 14 = 0.71. This area is approximately 0.4034.

Question 10

Use a table of areas to find the specified area under the standard normal curve.
-The area that lies between -0.73 and 2.27 
A)0.4884
B)0.7557
C)1.54
D)0.2211

Accepted Answer

The area under the standard normal curve between two z-scores (in this case, -0.73 and 2.27) is found by calculating the area from the mean to each z-score and then adding or subtracting these areas as appropriate. Looking up the areas corresponding to z-scores of -0.73 and 2.27 in a standard normal distribution table, we find the areas to the left of these z-scores. The area to the left of 2.27 is approximately 0.9883 and the area to the left of -0.73 is approximately 0.2327. Subtracting these (0.9883 - 0.2327) gives us the area between these two z-scores, which is 0.7556, rounded to 0.7557.

Question 11

Find the indicated probability or percentage for the normally distributed variable.
-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)0.9332
C)0.0668
D)0.5

Accepted Answer

The probability of a rating between 200 and 275 can be found using the standard normal distribution. First, calculate the z-scores for 200 and 275. The z-score for 200 is 0, and for 275 is (275-200)/50 = 1.5. Using the standard normal distribution table, the probability for z = 1.5 is approximately 0.9332, and for z = 0 is 0.5. The probability of being between these two z-scores is 0.9332 - 0.5 = 0.4332.

Question 12

Use a table of areas to obtain the shaded area under the standard normal curve.
-  
A)$0.1188$
B) $0.9406$
C) $0.0594$
D) $0.8812$

Accepted Answer

The answer of Use a table of areas to obtain...

Question 13

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution.
-Two percent of hair dryers produced in a certain plant are defective. Estimate the probability that of 10,000 randomly selected hair dryers, exactly 225 are defective. 
A)0.0065
B)0.0034
C)0.0051
D)0.0057

Accepted Answer

The answer of Estimate the indicated probability by using the...

Question 14

Find the indicated probability or percentage for the normally distributed variable.
-The variable $X$ is normally distributed. The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$. Find $P ( X > 16.1 )$.  
A)$0.8413$
B) $0.1550$
C) $0.1357$
D) $0.1587$

Accepted Answer

The answer of Find the indicated probability or percentage for...

Question 15

Use a table of areas to obtain the shaded area under the standard normal curve.
-  
A)$0.9896$
B) $0.0208$
C) $0.0104$
D) $0.9792$

Accepted Answer

The answer of Use a table of areas to obtain...

Question 16

Use a table of areas for the standard normal curve to find the required z-score.
-Determine the two z-scores that divide the area under the standard normal curve into a middle 0.874 area and two outside 0.063 areas. 
A)-1.45, 1.45
B)-1.46, 1.46
C)-1.53, 1.53
D)-1.39, 1.39

Accepted Answer

The answer of Use a table of areas for the...

Question 17

Use the empirical rule to solve the problem.
-At one college, GPAs are normally distributed with a mean of 2.8 and a standard deviation of 0.6. What percentage of students at the college have a GPA between 2.2 and 3.4? 
A)99.74%
B)95.44%
C)68.26%
D)84.13%

Accepted Answer

The empirical rule, or 68-95-99.7 rule, states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. Here, 2.2 and 3.4 are one standard deviation below and above the mean (2.8 ± 0.6), respectively. Therefore, about 68% of students have GPAs in this range.

Question 18

Use a table of areas to obtain the shaded area under the standard normal curve.
-  
A)$0.0183$
B) $0.0366$
C) $0.9634$
D) $0.9817$

Accepted Answer

The answer of Use a table of areas to obtain...

Question 19

Find the specified percentile, quartile, or decile.
-The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.2 and a standard deviation of 40.8. All units are in mg/100 mL. Find the 9th percentile. 
A)161.5 mg/100mL
B)123.5 mg/100mL
C)107.2 mg/100mL
D)165.1 mg/100mL

Accepted Answer

To find the 9th percentile of a normally distributed set of data, we use the formula for the z-score in relation to percentiles. The 9th percentile corresponds to a z-score that we can find using a z-table or a calculator designed for statistics. The formula to convert a z-score to the actual value in the context of a normal distribution is X = μ + Zσ, where X is the value we're looking for, μ is the mean, Z is the z-score corresponding to the 9th percentile, and σ is the standard deviation.For the 9th percentile, the z-score is approximately -1.34 (this value can be found in z-tables or statistical calculators). Using the given mean (μ = 178.2) and standard deviation (σ = 40.8), we can calculate the 9th percentile as follows:X = μ + Zσ = 178.2 + (-1.34)(40.8) ≈ 178.2 - 54.7 ≈ 123.5 mg/100 mL.Therefore, the 9th percentile of the serum cholesterol levels is approximately 123.5 mg/100 mL.

Question 20

Use a table of areas to obtain the shaded area under the standard normal curve.
-  
A)$0.8106$
B) $0.1894$
C) $0.6212$
D) $0.3788$

Accepted Answer

The answer of Use a table of areas to obtain...

Quiz 6: The Normal Distribution

Provide an appropriate response. Assume that the variable under consideration has a density curve. -Given that 40.5% of all possible observations of the variable are less than 12, determine the area under the density curve that lies to the right of 12.

Use the empirical rule to solve the problem. -The systolic blood pressure of 18-year-old women is normally distributed with a mean of 120 mmHg and a standard deviation of 12 mmHg. What percentage of 18-year-old women have a Systolic blood pressure between 96 mmHg and 144 mmHg?

Use a table of areas to find the specified area under the standard normal curve. -The area that lies between -1.10 and -0.36

Use a table of areas for the standard normal curve to find the required z-score. -Find the z-score for which the area under the standard normal curve to its left is 0.96

Use a table of areas to obtain the shaded area under the standard normal curve. -

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. -Estimate the probability of getting exactly 43 boys in 90 births.

Find the specified percentile, quartile, or decile. -Scores on an English test are normally distributed with a mean of 32.7 and a standard deviation of 8.2. Find the 41st percentile.

Use a table of areas to find the specified area under the standard normal curve. -The area that lies between -0.73 and 2.27

Use a table of areas to obtain the shaded area under the standard normal curve. -

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. -Two percent of hair dryers produced in a certain plant are defective. Estimate the probability that of 10,000 randomly selected hair dryers, exactly 225 are defective.

Find the indicated probability or percentage for the normally distributed variable. -The variable $X$ is normally distributed. The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find $P ( X > 16.1 )$ .

Use a table of areas to obtain the shaded area under the standard normal curve. -

Use a table of areas for the standard normal curve to find the required z-score. -Determine the two z-scores that divide the area under the standard normal curve into a middle 0.874 area and two outside 0.063 areas.

Use the empirical rule to solve the problem. -At one college, GPAs are normally distributed with a mean of 2.8 and a standard deviation of 0.6. What percentage of students at the college have a GPA between 2.2 and 3.4?

Use a table of areas to obtain the shaded area under the standard normal curve. -

Find the specified percentile, quartile, or decile. -The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.2 and a standard deviation of 40.8. All units are in mg/100 mL. Find the 9th percentile.

Use a table of areas to obtain the shaded area under the standard normal curve. -

Quiz 6: The Normal Distribution

Provide an appropriate response. Assume that the variable under consideration has a density curve. -Given that 40.5% of all possible observations of the variable are less than 12, determine the area under the density curve that lies to the right of 12.

Use the empirical rule to solve the problem. -The systolic blood pressure of 18-year-old women is normally distributed with a mean of 120 mmHg and a standard deviation of 12 mmHg. What percentage of 18-year-old women have a Systolic blood pressure between 96 mmHg and 144 mmHg?

Use a table of areas to find the specified area under the standard normal curve. -The area that lies between -1.10 and -0.36

Use a table of areas for the standard normal curve to find the required z-score. -Find the z-score for which the area under the standard normal curve to its left is 0.96

Use a table of areas to obtain the shaded area under the standard normal curve. -

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. -Estimate the probability of getting exactly 43 boys in 90 births.

Find the specified percentile, quartile, or decile. -Scores on an English test are normally distributed with a mean of 32.7 and a standard deviation of 8.2. Find the 41st percentile.

Use a table of areas to find the specified area under the standard normal curve. -The area that lies between -0.73 and 2.27

Use a table of areas to obtain the shaded area under the standard normal curve. -

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. -Two percent of hair dryers produced in a certain plant are defective. Estimate the probability that of 10,000 randomly selected hair dryers, exactly 225 are defective.

Find the indicated probability or percentage for the normally distributed variable. -The variable XXX is normally distributed. The mean is μ=15.2\mu = 15.2μ=15.2 and the standard deviation is σ=0.9\sigma = 0.9σ=0.9 . Find P(X>16.1) P ( X > 16.1 ) P(X>16.1) .

Use a table of areas to obtain the shaded area under the standard normal curve. -

Use a table of areas for the standard normal curve to find the required z-score. -Determine the two z-scores that divide the area under the standard normal curve into a middle 0.874 area and two outside 0.063 areas.

Use the empirical rule to solve the problem. -At one college, GPAs are normally distributed with a mean of 2.8 and a standard deviation of 0.6. What percentage of students at the college have a GPA between 2.2 and 3.4?

Use a table of areas to obtain the shaded area under the standard normal curve. -

Find the specified percentile, quartile, or decile. -The serum cholesterol levels for men in one age group are normally distributed with a mean of 178.2 and a standard deviation of 40.8. All units are in mg/100 mL. Find the 9th percentile.

Use a table of areas to obtain the shaded area under the standard normal curve. -

Find the indicated probability or percentage for the normally distributed variable. -The variable $X$ is normally distributed. The mean is $\mu = 15.2$ and the standard deviation is $\sigma = 0.9$ . Find $P ( X > 16.1 )$ .