Question

Chebyshev's Theorem and the Empirical Rule 1. For a distribution with mean 80 and standard deviation 10 A) What percentage of values will fall between 60 and 100? B) What percentage of values will fall between 50 and 110? 2. The average U.S. yearly per capita consumption of citrus fruit is 26.8 pounds. Suppose that the A) What percentage of Americans would you expect to consume more than 31 pounds of citrus d is bell-shaped with a standard deviation of 4.2 pounds. fruit per year? B) What percentage of Americans would you expect to consume less than 18.4 pounds of citrus fruit per year? C) What percentage of Americans would you expect to consume between 18.4 and 31 pounds of citrus fruit per year? 3. The average number of trials it took a sample of mice to learn to traverse a maze was 12. The standard deviation was 4. What percentage of data values will fall in the range of 4 to 20 trials? 4. The average score on a special test of the knowledge of wood finishing has a mean of 53 and a standard deviation of 6, Find the range of values in which at least 75% of the scores will lie. 5. The average full-time faculty member in a post-secondary degree-granting institution works an average of 53 hours per week. Assume that the distribution is bell-shaped and has a standard deviation of 2.8 hours. A) What percentage of fecuily members work more tham 58.6 hours a week? B) What percentage of faculty members work between 50.2 and 55.8 hours a week?

Answer 1

1)A) 80 - 2 * 10 = 60

        80 + 2 * 10 = 100

So 60 and 100 are two standard deviations from the mean.

P(60 < X < 100) = 1 - 1/k²

                           = 1 - 1/2²

                           = 1 - 1/4 = 3/4 = 0.75 = 75%

B) 80 - 3 * 10 = 50

    80 + 3 * 10 = 110

So 50 and 110 are three standard deviation from the mean.

So P(50 < X < 110) = 1 - 1/k²

                                = 1 - 1/3²

                                = 1 - 1/9 = 8/9 = 0.8889 = 88.89%

2)A) 26.8 + 4.2 = 31

31 is one standard deviation above the mean.

According to empirical rule 68% of the data fall within one standard deviation from the mean.

So 34% of the data fall between 26.8 and 31.

So 50% - 34% = 16% Americans would expect to consume more than 31 pounds of citrus fruit per year.

B) 26.8 - 2 * 4.2 = 18.4

18.4 is two standard deviation below the mean.

According to empirical rule 95% of the data fall within two standard deviation from the mean.

So 47.5% of the data fall between 18.4 and 26.8.

So 50% - 47.5% = 2.5% Americans would expect to consume less than 18.4 pounds of citrus fruit per year.

C) 34% of the data fall between 26.8 and 31 and 47.5% of the data fall between 18.4 and 26.8.

So total 34% + 47.5% = 81.5% Americans would expect to consume between 18.4 and 31 pounds of citrus fruit per year.

3) 12 - 2 * 4 = 12 - 8 = 4

     12 + 2 * 4 = 12 + 8 = 20

So 4 and 20 are two standard deviations from the mean.

P(4 < X < 20) = 1 - 1/k²

                        = 1 - 1/2²

                        = 1 - 1/4 = 3/4 = 0.75 = 75%

4) According to Chebyshev's theorem 75% of the data falls within two standard deviation from the mean.

53 - 2 * 6 = 53 - 12 = 41

53 + 2 * 6 = 53 + 12 = 65

So 75% of the scores will lie between the range 41 and 65.

5)A) 53 + 2 * 2.8 = 58.6

58.6 is two standard deviation above the mean.

According to empirical rule 95% of the data fall within two standard deviation from the mean.

So 47.5% of the data fall between 53 and 58.6.

So 50% - 47.5% = 2.5% of faculty members work more than 58.6 hours a week.

B) 53 - 2.8 = 50.2

    53 + 2.8 = 55.8

According to empirical rule 68% of the data fall within one standard deviation from the mean.

So 68% of faculty members work between 50.2 and 55.8 hours a week.