Question

In: Statistics and Probability

1. Suppose a car driven under specific conditions gets a mean gas mileage of 40 miles...

1. Suppose a car driven under specific conditions gets a mean gas mileage of 40 miles per gallon with a standard deviation of 3 miles per gallon. On about what percentage of the trips will your gas mileage be above 43 miles per gallon?

A.About 68%, because 43 miles per gallon is 2 std. deviations above the mean. By the 68-95-99.7 rule, about 68% of the distribution lies within 2 std. deviations of the mean.

B.About 68%, because 43 miles per gallon is 1 std. deviation above the mean. By the 68-95-99.7 rule, about 68% of the distribution lies within 1 std. deviation of the mean.

C.About 16%, because 43 miles per gallon is 1 std. deviation above the mean. By the 68-95-99.7 rule, about 16% of the distribution lies within 1 std. deviation of the mean.

D.About 2.5%, because 43 miles per gallon is 2 std. deviations above the mean. By the 68-95-99.7 rule, about 2.5% of the distribution lies within 2 std. deviations of the mean.

E.About 16%, because 43 miles per gallon is 1 std. deviation above the mean. By the 68-95-99.7 rule, about 68% of the distribution lies within 1 std. deviation of the mean. So 32% lies outside of this range, 16% in each tail.

F.About 2.5%, because 43 miles per gallon is 2 std. deviations above the mean. By the 68-95-99.7 rule, about 95% of the distribution lies within 2 std. deviations of the mean. So 5% lies outside of this range, 2.5% in each tail.

2.Assume that a set of test scores is normally distributed with a mean of 80 and a standard deviation of 25. Use the 68-95-99.7 rule to find the following quantities.

a. The percentage of scores less than 80 is ___________%. (Round to one decimal place as needed.)

b. The percentage of scores greater than 105 is _____________%. (Round to one decimal place as needed.)

c. The percentage of scores between 30 and 105 is ___________%.(Round to one decimal place as needed.)

3.Working with equal sample sizes, researchers compared three new cold remedies to a placebo. Which remedy is most likely to be most effective? Explain.

A.The one that gave results statistically significant at the 0.01 level, because the remedy was successful in so many people that the chance of at least this much success would be greater than 0.01, if the remedy were no better than the placebo.

B.The one that gave results that were not statistically significant, because the other tests do not produce positive results.

C.The one that gave results that were not statistically significant, because statistically significant results would indicate that the test was not performed correctly.

D.The one that gave results statistically significant at the 0.05 level, because the remedy was successful in so many people that the chance of at least this much success would be greater than 0.05, if the remedy were no better than the placebo.

E.The one that gave results statistically significant at the 0.01 level, because the remedy was successful in so many people that the chance of at least this much success would be less than 0.01, if the remedy were no better than the placebo.

F.The one that gave results statistically significant at the 0.05 level, because the remedy was successful in so many people that the chance of at least this much success would be less than 0.05, if the remedy were no better than the placebo.

Solutions

Expert Solution

1.

Mean, =40

Standard deviation, =3

43 =40+3 =​​​​​​. So, above 43 is one std.deviation above the mean.

Thus, Option E. is correct.

E. About 16%, because 43 miles per gallon is 1 std. deviation above the mean. By the 68-95-99.7 rule, about 68% of the distribution lies within 1 std. deviation of the mean. So 32% lies outside of this range, 16% in each tail.

​​​​​​

2.

Mean, =80

Standard deviation, =25

a.

P(X < 80) =0.50 =50%

(because the area is equal on both sides of the mean, i.e., 0.50 and 0.50).

b.

P(X > 105) =P(X > (80+(1*25)) =(1 - 0.68)/2 =0.32/2 =0.16 =16%

c.

P(30 X 105) =P(80 - (2*25) X 80+(1*25)) =(0.95/2)+(0.68/2) =0.475 + 0.34 =0.815 =81.5%

3.

Option E. is correct.

E. The one that gave results statistically significant at the 0.01 level, because the remedy was successful in so many people that the chance of at least this much success would be less than 0.01, if the remedy were no better than the placebo.


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