Question

The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of fresh fruit per person is 99.9 pounds. The standard deviation of fresh fruit consumption is about 30 pounds. Suppose a researcher took a random sample of 38 people and had them keep a record of the fresh fruit they ate for one year. Appendix A Statistical Tables (Round all z values to 2 decimal places. Round your answers to 4 decimal places.) a. What is the probability that the sample average would be less than 90 pounds? p = b. What is the probability that the sample average would be between 98 and 105 pounds? p = c. What is the probability that the sample average would be less than 112 pounds? p = d. What is the probability that the sample average would be between 93 and 96 pounds? p =

Answer 1

Let M be the sample average $\overline{X}$ .

Thus, E(M) = 99.9, s.d.(M) = 30 / $\sqrt{38}$ = 4.8666.

Thus, M ~ N(99.9, 4.8666) i.e. (M - 99.9)/4.8666 ~ N(0,1).

a) The required probability = P(M < 90)

= P[(M - 99.9)/4.8666 < (90 - 99.9)/4.8666]

= P[(M - 99.9)/4.8666 < 2.0343] = $\Phi$ (2.0343) = 0.979. (Ans).

[ $\Phi$ (.) is the cdf of N(0,1)].

b) P(98 < M < 105)

= P[(98 - 99.9)/4.8666 < (M - 99.9)/4.8666 < (105 - 99.9)/4.8666]

= P[-0.3904 < (M - 99.9)/4.8666 < 0.2160]

= $\Phi$ (0.2160) - $\Phi$ (-0.3904) = 0.5855 - 0.3481 = 0.2374. (Ans).

c) P(M < 112) = P[(M - 99.9)/4.8666 < (112 - 99.9)/4.8666]

= P[(M - 99.9)/4.8666 < 2.4863] = $\Phi$ (2.4863) = 0.9935. (Ans).

d) P(93 < M < 96)

= P[(93 - 99.9)/4.8666 < (M - 99.9)/4.8666 < (96 - 99.9)/4.8666]

= P[ -1.4178 < (M - 99.9)/4.8666 < -0.8014]

= $\Phi$ (-0.8014) - $\Phi$ (-1.4178) = 0.2115 - 0.0781 = 0.1334. (Ans).