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Let y be a random variable. In a population, ay = 119 and 62 = 54. Use the central limit theorem to answer the following ques

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Answer #1

Solution:

Population mean, muy = 119

Variance, sigma2 = 54

Central limit theorem, CLT says Z = (Y(bar) - muy)/(sigma2/n)1/2

Probability for any value of Y is found using z value and standard normal distribution table. Since, area under normal distribution for a z-value is of left side:

Pr(Y(bar) < Y) = Pr(z = Z), and

Pr(Y(bar) > Y) = 1 - Pr(z = Z)

Then,

a) With n = 100:

Z = (120 - 119)/(54/100)1/2 = 1.36

So, Pr(Y(bar) < 120) = Pr(z = 1.36) = 0.9131

b) With n = 72:

Z = (120 - 119)/(54/72)1/2 = 1.15

Z = (125 - 119)/(54/72)1/2 = 6.93

So, Pr(Y(bar) < 120) = Pr(z = 1.15) = 0.8749

Pr(Y(bar) < 125) = Pr(z = 6.93) = 1 (nearly)

So, Pr(120 < Y(bar) < 125) = 1 - 0.8749 = 0.1251

c) With n = 116:

Z = (121 - 119)/(54/116)1/2 = 2.93

So, Pr(Y(bar) > 121) = 1 - Pr(z = 2.93)

Pr(Y(bar) > 121) = 1 - 0.9983 = 0.0017

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