Question

In: Statistics and Probability

Given a population with a mean of µ = 100 and a variance σ2 = 14,...

Given a population with a mean of µ = 100 and a variance σ2 = 14, assume the central limit theorem applies when the sample size is n ≥ 25. A random sample of size n = 46 is obtained. What is the probability that 98.80 < x¯ < 100.92?

Solutions

Expert Solution

Solution :

Given that,

mean = = 100

standard deviation = = 3.7417

n = 46

= = 100

= / n = 3.7417/ 46 = 0.5517

P( 98.80 < < 100.92) = P((98.80 - 100) / 0.5517<( - ) / < (100.92 - 100) / 0.5517))

= P(-2.18 < Z < 1.67)

= P(Z < 1.67) - P(Z < -2.18) Using standard normal table,  

= 0.9525 - 0.0146

= 0.9379

Probability = 0.9379


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