Question

(15.12) The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths trapped is only 0.6 , but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard deviation 0.8 . What is the mean (±0.1) of the average number of moths x¯¯¯x¯ in 50 traps? ___ And the standard deviation of the average number of moths? (±0.001) ___ Use the central limit theorem to approximate the probability (±0.01) that the average number of moths in 50 traps is greater than 0.5 : ____

(15.28) Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score μμ of those who took the MCAT on your campus, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 6.5. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 26. In answering the following, use z-scores rounded to two decimal places. If you choose one student at random, what is the probability (±±0.0001) that the student's score is between 20 and 30? ___ You sample 21 students. What is the standard deviation (±±0.01) of sampling distribution of their average score x¯¯¯x¯? ____ What is the probability (±±0.0001) that the mean score of your sample is between 20 and 30? ____

Answer 1

Note: P(z>a) = 1-P(z<a)

Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(-0.8839, TRUE)" to find the probability.

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Note: P(a<z<b) = P(z<b)-P(z<a)

Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(0.6154, TRUE)" & "=NORM.S.DIST(-0.9231, TRUE)" to find the probability.

Note: P(a<z<b) = P(z<b)-P(z<a)

Refer Standard normal table/Z-table to find the probability or use excel formula "=NORM.S.DIST(2.8200, TRUE)" & "=NORM.S.DIST(-4.2301, TRUE)" to find the probability.