In: Statistics and Probability
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?
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