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?

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Expert Solution

Solution :

Given that,

mean = = 100

standard deviation = = 3.6056

n = 28

= = 100

= / n = 3.6056 / 28 = 0.6814

P(98.02 < < 99.08) = P((98.2 - 100) /0.6814 <( - ) / < (99.08 - 100) / 0.6814))

= P(-2.91 < Z <-1.35)

= P(Z < -1.35) - P(Z < -2.91) Using z table,

= 0.0885 - 0.0018

= 0.0867

The probability is 0.0867

orchestra answered 1 year ago

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