Question

# Given a population with a mean of µ = 100 and a variance σ2 = 13,...

Given a population with a mean of µ = 100 and a variance σ2 = 13, assume the central limit theorem applies when the sample size is n ≥ 25. A random sample of size n = 28 is obtained. What is the probability that 98.02 < x⎯⎯ < 99.08?

#### Homework Answers

Answer #1

Using central limit theorem,

P($\bar{x}$ < x) = P (Z < x - $\mu$ / $\sigma$ / sqrt(n) )

So,

P( 98.02 < $\bar{x}$ < 99.08) = P( $\bar{x}$ < 99.08) - P( $\bar{x}$ < 98.02)

= P( Z < 99.08 - 100 / sqrt(13) / sqrt(28) ) - P( Z < 98.02 - 100 / sqrt(13) / sqrt(28) )

= P (Z < -1.3502) - P( Z < -2.9058)

= 0.0885 - 0.0018 (From Z table)

= 0.0867

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