Question 1

Find the indicated percentage for the normally distributed variable. Round your answer to two decimal places, if necessary.
-A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and astandard deviation of 50. What percentage of the ratings will be between 200 and 300?

Accepted Answer

The z-score for 300 is (300-200)/50 = 2. The percentage of data between the mean (z=0) and z=2 in a standard normal distribution is approximately 47.72%.

Question 2

Find the indicated percentage for the normally distributed variable. Round your answer to two decimal places, if necessary.
-The volumes of soda in quart soda bottles are normally distributed with a mean of 32 ounces and a standard deviation of 1.2 ounces. What percentage of soda bottles will have a volume less than 31.46 ounces?

Accepted Answer

To find the percentage of soda bottles with a volume less than 31.46 ounces, we use the Z-score formula: Z = (X - μ) / σ, where X is the value (31.46 ounces), μ is the mean (32 ounces), and σ is the standard deviation (1.2 ounces). Plugging in the values, we get Z = (31.46 - 32) / 1.2 = -0.45. Looking up a Z-score of -0.45 in a standard normal distribution table or using a calculator, we find that the percentage of values below this Z-score is approximately 32.64%.

Question 3

Apply the 68-95-99.7 rule to answer the question.
-The annual precipitation for one city is normally distributed with a mean of 266 inches and a standard deviationof 2.5 inches. What percentage of years had precipitation between 261 inches and 271 inches?

Accepted Answer

The 68-95-99.7 rule, also known as 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. The range from 261 inches to 271 inches encompasses a span of two standard deviations (2.5 inches x 2 = 5 inches) from the mean (266 inches), so approximately 95% of the years had precipitation within this range.

Question 4

Find the indicated percentage for the normally distributed variable. Round your answer to two decimal places, if necessary.
-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 human pregnancy lasts at least 289 days?

Accepted Answer

To find the probability that a human pregnancy lasts at least 289 days, we use the Z-score formula: Z = (X - μ) / σ, where X is the value of interest (289 days), μ is the mean (268 days), and σ is the standard deviation (15 days). Calculating the Z-score gives us Z = (289 - 268) / 15 = 21 / 15 = 1.4. Looking up the Z-score of 1.4 in a standard normal distribution table or using a calculator, we find that the area to the left of Z = 1.4 is approximately 0.9192, which means the area to the right (the probability of a pregnancy lasting at least 289 days) is 1 - 0.9192 = 0.0808 or 8.08%.

Question 5

Find the requested mean or standard deviation.
-The heights of a large population of students have a mean of 60" with a standard deviation of 4". What is the mean of the resulting distribution of sample means for n =16?

Accepted Answer

The mean of the resulting distribution of sample means (also known as the sampling distribution of the sample mean) is equal to the mean of the population, which is 60".

Question 6

Solve the problem.
-The area under the standard normal distribution curve between 1 and 2 is equal to 0.1359. Scores on a particular aptitude test are normally distributed with a mean of 100 and a standard deviation of 10. Which of the following are equal to 13.59%?
A. The percentage of scores between 120 and 130
B. The percentage of scores between 110 and 120
C. The percentage of scores between 80 and 90
D. The percentage of scores between 90 and 120
E. The percentage of scores between 90 and 110

Accepted Answer

The answer of Solve the problem.
-The area under the standard...

Question 7

Find the requested percentile.
-At one college, GPA's are normally distributed with a mean of 2.6 and a standard deviation of 0.6. Find the 75th percentile. Round your answer to one decimal place.

Accepted Answer

The answer of Find the requested percentile.
-At one college, GPA's...

Question 8

Find the requested percentile.
-One college requires that scholarship students maintain a GPA of 3.3, when the GPAs at that college are normally distributed with a mean of 3.0 and a standard deviation of 0.6. What percentile must scholarship students
Remain in?

Accepted Answer

To find the percentile, we calculate the z-score: z = (3.3 - 3.0) / 0.6 = 0.5. Using a standard normal distribution table, a z-score of 0.5 corresponds to approximately the 69th percentile.

Question 9

Apply the 68-95-99.7 rule to answer the question.
-At one college, GPA's are normally distributed with a mean of 2.8 and a standard deviation of 0.5. What percentage of students at the college have a GPA between 2.3 and 3.3?

Accepted Answer

The 68-95-99.7 rule states that approximately 68% of data falls within one standard deviation of the mean in a normal distribution. Since 2.3 and 3.3 are one standard deviation below and above the mean (2.8 ± 0.5), about 68% of students have a GPA between 2.3 and 3.3.

Question 10

Find the indicated percentage for the normally distributed variable. Round your answer to two decimal places, if necessary.
-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?

Accepted Answer

The z-score for a diameter of 0.32 inches is (0.32 - 0.30) / 0.01 = 2. Using a standard normal distribution table, we find that the percentage of bolts with a diameter greater than 0.32 inches is approximately 2.28%.

Question 11

Find the indicated percentage for the normally distributed variable. Round your answer to two decimal places, if necessary.
-Assume that the weights of nickels are normally distributed with a mean of 5.000 grams and a standard deviation of 0.06 grams. A vending machine will reject all coins with weights more than 1 standard deviation above or below the mean. What percentage of legal nickels will be rejected by the machine?

Accepted Answer

In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. Therefore, if the vending machine rejects nickels more than 1 standard deviation from the mean, it will reject approximately 32% of nickels (100% - 68% = 32%).

Question 12

Apply the 68-95-99.7 rule to answer the question.
-The annual precipitation for one city is normally distributed with a mean of 35.4 inches and a standard deviation of 3.9 inches. What percentage of years had precipitation between 23.7 inches and 47.1 inches?

Accepted Answer

The 68-95-99.7 rule, also known as the empirical rule, states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations. The range from 23.7 inches to 47.1 inches encompasses the mean (35.4 inches) and extends beyond three standard deviations (3 x 3.9 = 11.7 inches) in both directions, covering virtually all possible outcomes (99.7%).

Question 13

Find the indicated percentage for the normally distributed variable. Round your answer to two decimal places, if necessary.
-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. What percentage of the ratings will be between 180 and 210?

Accepted Answer

The correct answer is A because the percentage of the ratings that will be between 180 and 210 is 23.47%. This can be calculated using a normal distribution calculator or by using the z-score formula.

Question 14

Find the requested percentile.
-Scores on an English test are normally distributed with a mean of 31.7 and a standard deviation of 6.6. Find the 41st percentile. Round your answer to one decimal place.

Accepted Answer

To find the 41st percentile, we use the z-score formula for a normal distribution. The z-score corresponding to the 41st percentile is approximately -0.23. Using the formula X = μ + zσ, where μ = 31.7 and σ = 6.6, we calculate X = 31.7 + (-0.23)(6.6) ≈ 30.2.

Question 15

Find the indicated percentage for the normally distributed variable. Round your answer to two decimal places, if necessary.
-The monthly incomes of trainees at a local mill are normally distributed, with a mean of $1,100 and a standard deviation $150. What percentage of trainees earn less than $875 a month?

Accepted Answer

To find the percentage of trainees earning less than $875, we calculate the z-score using the formula $z = \frac{X - \mu}{\sigma}$, where $X$ is the value in question ($875), $\mu$ is the mean ($1,100), and $\sigma$ is the standard deviation ($150). Plugging in the values, we get $z = \frac{875 - 1100}{150} = -1.5$. Looking up a z-score of -1.5 in a standard normal distribution table or using a calculator, we find that approximately 6.68% of the distribution falls below this z-score. This means about 6.68% of trainees earn less than $875 a month.

Question 16

Apply the 68-95-99.7 rule to answer the question.
-The amount of Jen's monthly phone bill is normally distributed with a mean of $50 and a standard deviation of $10. What percentage of her phone bills are between $40 and $60?

Accepted Answer

According to the 68-95-99.7 rule, 68% of data in a normal distribution falls within one standard deviation of the mean. Here, $40 to $60 is within one standard deviation ($10) from the mean ($50).

Question 17

Find the requested percentile.
-The weights of certain machine components are normally distributed with a mean of 8.62 grams and a standard deviation of 0.09 grams. Find the 97th percentile. Round your answer to one decimal place.

Accepted Answer

The 97th percentile corresponds to a z-score of approximately 1.88. Using the formula for a normal distribution (X = μ + zσ), where μ = 8.62 and σ = 0.09, the calculation is 8.62 + (1.88 * 0.09) = 8.7882, which rounds to 8.8 grams.

Question 18

Apply the 68-95-99.7 rule to answer the question.
-The amount of Jen's monthly phone bill is normally distributed with a mean of $78 and a standard deviation of $10. What percentage of her phone bills are between $48 and $108?

Accepted Answer

The 68-95-99.7 rule, also known as 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, $48 and $108 are three standard deviations below and above the mean ($78 ± 3*$10), respectively. Therefore, 99.7% of Jen's phone bills are between $48 and $108.

Question 19

Solve the problem.
-Which of the following statements concerning areas under the standard normal distribution curve is/are true (if any)?
A. The area to the left of -3 in a standard normal distribution curve is zero.
B. The area under a standard normal distribution curve between any two z-scores is greater thazn ero.
C. The area under the standard normal normal distribution curve between two z-scores will be negative if
Both z-scores are negative.
D. The area under the standard normal distribution curve to the left of any z-score is less than 1.

Accepted Answer

B: The area between any two z-scores under a standard normal distribution curve is always positive because it represents the probability of a value falling between those two points, and probabilities cannot be negative. D: The total area under the standard normal distribution curve is 1, so the area to the left of any z-score must be less than 1, as it represents the cumulative probability up to that z-score.

Question 20

Solve the problem.
-Which of the following statements concerning areas under the standard normal distribution curve is/are true?
A. If a z-score is negative, the area to its right is greater than 0.5
B. If the area to the right of a z-score is less than 0.5, the z-score is negative.
C. If a z-score is positive, the area to its left is less than 0.5

Accepted Answer

The answer of Solve the problem.
-Which of the following statements...

Quiz 5: A Normal World

Apply the 68-95-99.7 rule to answer the question. -The annual precipitation for one city is normally distributed with a mean of 266 inches and a standard deviationof 2.5 inches. What percentage of years had precipitation between 261 inches and 271 inches?

Find the requested mean or standard deviation. -The heights of a large population of students have a mean of 60" with a standard deviation of 4". What is the mean of the resulting distribution of sample means for n =16?

Find the requested percentile. -At one college, GPA's are normally distributed with a mean of 2.6 and a standard deviation of 0.6. Find the 75th percentile. Round your answer to one decimal place.

Find the requested percentile. -One college requires that scholarship students maintain a GPA of 3.3, when the GPAs at that college are normally distributed with a mean of 3.0 and a standard deviation of 0.6. What percentile must scholarship students Remain in?

Apply the 68-95-99.7 rule to answer the question. -At one college, GPA's are normally distributed with a mean of 2.8 and a standard deviation of 0.5. What percentage of students at the college have a GPA between 2.3 and 3.3?

Apply the 68-95-99.7 rule to answer the question. -The annual precipitation for one city is normally distributed with a mean of 35.4 inches and a standard deviation of 3.9 inches. What percentage of years had precipitation between 23.7 inches and 47.1 inches?

Find the requested percentile. -Scores on an English test are normally distributed with a mean of 31.7 and a standard deviation of 6.6. Find the 41st percentile. Round your answer to one decimal place.

Apply the 68-95-99.7 rule to answer the question. -The amount of Jen's monthly phone bill is normally distributed with a mean of $50 and a standard deviation of $10. What percentage of her phone bills are between $40 and $60?

Find the requested percentile. -The weights of certain machine components are normally distributed with a mean of 8.62 grams and a standard deviation of 0.09 grams. Find the 97th percentile. Round your answer to one decimal place.

Apply the 68-95-99.7 rule to answer the question. -The amount of Jen's monthly phone bill is normally distributed with a mean of $78 and a standard deviation of $10. What percentage of her phone bills are between $48 and $108?