Chapter 7: Problem 16 Previous Problem Problem List Next Problem (1 point) The d

Question

Chapter 7: Problem 16 Previous Problem Problem List Next Problem (1 point) The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 8.1 ounces and standard deviation 0.14 ounces. (a) What is the probability that the average weight of a bar in a Simple Random Sample (SRS) with 4 of these chocolate bars is between 7.91 and 8.25 ounces? ANSWER: (b) For a SRS of 4 of these chocolate bars, what is the level L such that there is a 5% chance that the average weight is less than L? ANSWER:

Explanation / Answer

Chapter 7 problem 16

(a) Mean weight = 8.1 ounces

Standard deviation of weight = 0.14 onces

Here sample size n = 4

standard error of sample mean se0 = /sqrt(n) = 0.14/2 = 0.07

Pr(7.91 < x < 8.25) = Pr(x < 8.25) - Pr(x< 7.91) = (Z2 ) - (Z1)

Z2 = (8.25 - 8.1) /0.07 = 2.1426

Z1 = (7.91 - 8.1)/ 0.07 = -2.7143

(Z2 ) - (Z1) = (2.1426) - (-2.7143) = 0.9839 - 0.0033 = 0.9806

(b) Pr(x > x0 ) = 0.05

Pr(x < x0) = 0.95

so respective Z - value

Z = -1.645

-1.645 = (x0 < x)/ 0.07

x0 = 8.10 - 0.07 * 1.645 = 7.9849 ounces = L

Chapter 7 : Problem 15

Mean claim amount = $ 680

Standard deviation of claim = $ 640

(a) Pr (x < $ 630)

Z = (630 - 680)/ 640 = -0.078125

Pr (x < $ 630) = Pr(Z < 0.078125) = 0.4689

(b) Here sample size = 70

standard error of sampel mean se0 = s/ sqrt(n) = 640/ sqrt(70) = 76.4946

Pr(x < $ 630 ; $ 680 ; $ 76.4946)

Z = (630 - 680)/ 76.4946 = -0.6536

Pr(x < $ 630 ; $ 680 ; $ 76.4946) = Pr(Z < -0.6536) = 0.2567

(c) There a sample larger than 70 claims is considerd , there is "LESS " chance

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