Quiz over average calculator!

1) Cucumbers grown on a certain farm have weights with a standard deviation of 2 ounces. What is the mean weight if 85% of the cucumbers weigh less than 16 ounces?

13.92

14.30

14.40

14.88

15.70

The mean weight of the cucumbers is 13.92 ounces. This can be determined by using a normal distribution table and finding the z-score associated with the 85th percentile (which is 1.036). Then, using the formula z = (x - μ) / σ, where z is the z-score, x is the value, μ is the mean, and σ is the standard deviation, we can rearrange the formula to solve for the mean. Plugging in the values, 1.036 = (16 - μ) / 2, we can solve for μ, which is 13.92.

Explanation

The mean weight of the cucumbers is 13.92 ounces. This can be determined by using a normal distribution table and finding the z-score associated with the 85th percentile (which is 1.036). Then, using the formula z = (x - μ) / σ, where z is the z-score, x is the value, μ is the mean, and σ is the standard deviation, we can rearrange the formula to solve for the mean. Plugging in the values, 1.036 = (16 - μ) / 2, we can solve for μ, which is 13.92.

2) A trucking firm determines that its fleet of trucks averages a mean of 12.4 miles per gallon with a standard deviation of 1.2 miles per gallon on cross-country hauls. What is the probability that one of the trucks averages fewer than 10 miles per gallon?

.0082

.0228

.4772

.5228

.9772

The question asks for the probability that one of the trucks averages fewer than 10 miles per gallon. To find this probability, we need to calculate the z-score for 10 miles per gallon using the formula z = (x - μ) / σ, where x is the value we are interested in (10), μ is the mean (12.4), and σ is the standard deviation (1.2).

Calculating the z-score, we have z = (10 - 12.4) / 1.2 = -2.

Next, we look up the area to the left of the z-score -2 in the standard normal distribution table. The area is approximately 0.0228, which corresponds to the probability that one of the trucks averages fewer than 10 miles per gallon. Therefore, the correct answer is 0.0228.

Explanation

The question asks for the probability that one of the trucks averages fewer than 10 miles per gallon. To find this probability, we need to calculate the z-score for 10 miles per gallon using the formula z = (x - μ) / σ, where x is the value we are interested in (10), μ is the mean (12.4), and σ is the standard deviation (1.2).

Calculating the z-score, we have z = (10 - 12.4) / 1.2 = -2.

Next, we look up the area to the left of the z-score -2 in the standard normal distribution table. The area is approximately 0.0228, which corresponds to the probability that one of the trucks averages fewer than 10 miles per gallon. Therefore, the correct answer is 0.0228.

3) A random survey was conducted to determine the cost of residential gas heat. Analysis of the survey results indicated that the mean monthly cost of gas was $125, with a standard deviation of $10. What is the z-score of a house with a gas bill of $150 per month?

-2.5

0.40

1.5

2.5

25

The z-score measures how many standard deviations a particular value is from the mean. In this case, the mean monthly cost of gas is $125 with a standard deviation of $10. To find the z-score of a house with a gas bill of $150 per month, we need to calculate how many standard deviations $150 is from the mean.

Z-score = (Value - Mean) / Standard Deviation

Z-score = ($150 - $125) / $10 = 25 / $10 = 2.5

Therefore, the z-score of a house with a gas bill of $150 per month is 2.5.

Explanation

The z-score measures how many standard deviations a particular value is from the mean. In this case, the mean monthly cost of gas is $125 with a standard deviation of $10. To find the z-score of a house with a gas bill of $150 per month, we need to calculate how many standard deviations $150 is from the mean.

Z-score = (Value - Mean) / Standard Deviation

Z-score = ($150 - $125) / $10 = 25 / $10 = 2.5

Therefore, the z-score of a house with a gas bill of $150 per month is 2.5.

4) The average noise level in a restaurant is 30 decibels with a standard deviation of 4 decibels. Ninety-nine percent of the time it is below what value?

20.7

32.0

33.4

37.8

39.3

The given question asks for the value below which the noise level in the restaurant is 99% of the time. Since the average noise level is 30 decibels and the standard deviation is 4 decibels, we can use the z-score formula to find the value. The z-score corresponding to 99% is approximately 2.33. Multiplying this z-score by the standard deviation and adding it to the mean gives us 39.3, which is the value below which the noise level is 99% of the time.

Explanation

The given question asks for the value below which the noise level in the restaurant is 99% of the time. Since the average noise level is 30 decibels and the standard deviation is 4 decibels, we can use the z-score formula to find the value. The z-score corresponding to 99% is approximately 2.33. Multiplying this z-score by the standard deviation and adding it to the mean gives us 39.3, which is the value below which the noise level is 99% of the time.

5) Which of the following are true statements? I. The area under the standard normal curve between 0 and 2 is twice the area between 0 and 1. II. The area under the standard normal curve between 0 and 2 is half the area between -2 and 2. III. For the standard normal curve, the interquartile range is approximately 3.

I and II

I and III

II and III

I, II and III

None of the above gives the complete set of true responses.

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Explanation

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6) A symmetric, mound-shaped distribution has a mean of 42 and a standard deviation of 7. Which of the following is true?

There are more data values between 42 and 49 than between 28 and 35.

The maximum value is less than 70.

Approximately 95% of the data lie between 35 and 49.

The interquartile range is approximately 14.

None of these is true.

In a symmetric, mound-shaped distribution, the mean is located at the center of the distribution. Since the mean is 42, it is reasonable to expect that there would be more data values between 42 and 49, which is slightly above the mean, compared to between 28 and 35, which is slightly below the mean. Therefore, the statement "There are more data values between 42 and 49 than between 28 and 35" is true.

Explanation

In a symmetric, mound-shaped distribution, the mean is located at the center of the distribution. Since the mean is 42, it is reasonable to expect that there would be more data values between 42 and 49, which is slightly above the mean, compared to between 28 and 35, which is slightly below the mean. Therefore, the statement "There are more data values between 42 and 49 than between 28 and 35" is true.

7) Populations P1 and P2 are normally distributed and have identical means. However, the standard deviation of P1 is twice the standard deviation of P2. What can be said about the percentage of observations falling within two standard deviations of the mean for each population?

The percentage for P1 is twice the percentage for P2.

The percentage for P1 is greater, but not twice as great, as the percentage for P2.

The percentage for P2 is twice the percentage for P1.

The percentage for P2 is greater, but not twice as great, as the percentage for P1.

The percentages are identical.

If two populations, P1 and P2, have identical means but different standard deviations, it means that the spread of data in P1 is greater than in P2. However, the question specifically asks about the percentage of observations falling within two standard deviations of the mean for each population. Since both populations have the same mean, the percentage of observations falling within two standard deviations of the mean will be the same for both populations. Therefore, the correct answer is that the percentages are identical.

Explanation

If two populations, P1 and P2, have identical means but different standard deviations, it means that the spread of data in P1 is greater than in P2. However, the question specifically asks about the percentage of observations falling within two standard deviations of the mean for each population. Since both populations have the same mean, the percentage of observations falling within two standard deviations of the mean will be the same for both populations. Therefore, the correct answer is that the percentages are identical.

8) The mean score on a college placement exam is 500 with a standard deviation of 100. Ninety-five percent of the test takers score above what?

260

336

405

414

664

The mean score on the college placement exam is 500 with a standard deviation of 100. This means that the majority of test takers will score around 500, with a spread of scores around that mean. Since we are looking for the score that 95% of test takers score above, we need to find the z-score corresponding to the 95th percentile. Using a standard normal distribution table, we find that the z-score corresponding to the 95th percentile is approximately 1.645. We can then use the formula z = (x - μ) / σ to solve for x, where x is the score we are looking for. Plugging in the values, we get 1.645 = (x - 500) / 100. Solving for x, we find that x is approximately 336. Therefore, 95% of test takers score above 336.

Explanation

The mean score on the college placement exam is 500 with a standard deviation of 100. This means that the majority of test takers will score around 500, with a spread of scores around that mean. Since we are looking for the score that 95% of test takers score above, we need to find the z-score corresponding to the 95th percentile. Using a standard normal distribution table, we find that the z-score corresponding to the 95th percentile is approximately 1.645. We can then use the formula z = (x - μ) / σ to solve for x, where x is the score we are looking for. Plugging in the values, we get 1.645 = (x - 500) / 100. Solving for x, we find that x is approximately 336. Therefore, 95% of test takers score above 336.

9) The average life expectancy of males in a particular town is 75 years, with a standard deviation of 5 years. Assuming that the distribution is mound-shaped and symmetric, the approximate 15th percentile in the age distribution is

60

65

70

75

80

The 15th percentile represents the value below which 15% of the data falls. In a mound-shaped and symmetric distribution, the mean and median are equal. Since the average life expectancy is 75 years, it is reasonable to assume that the median is also around 75 years. Therefore, the approximate 15th percentile would be slightly lower than the median, which is around 70 years.

Explanation

The 15th percentile represents the value below which 15% of the data falls. In a mound-shaped and symmetric distribution, the mean and median are equal. Since the average life expectancy is 75 years, it is reasonable to assume that the median is also around 75 years. Therefore, the approximate 15th percentile would be slightly lower than the median, which is around 70 years.

10) Assuming that batting averages have a bell-shaped distribution, arrange these in ascending order: I. An average with a z-score of -1. II. An average with a percentile rank of 20%. III. An average at the first quartile.

I, II, III

III, I, II

II, I, III

II, III, I

III, II, I

The z-score measures the number of standard deviations a data point is from the mean. A z-score of -1 indicates that the average is one standard deviation below the mean. The percentile rank of 20% means that the average is at the 20th percentile, which is below the median. The first quartile represents the 25th percentile, so it is lower than both the average with a z-score of -1 and the average with a percentile rank of 20%. Therefore, the correct order is I, II, III.

Explanation

The z-score measures the number of standard deviations a data point is from the mean. A z-score of -1 indicates that the average is one standard deviation below the mean. The percentile rank of 20% means that the average is at the 20th percentile, which is below the median. The first quartile represents the 25th percentile, so it is lower than both the average with a z-score of -1 and the average with a percentile rank of 20%. Therefore, the correct order is I, II, III.

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