41Sampling distribution of mean

A population has mean 20 and standard deviation 6. For a sample of 36, use $\displaystyle \Phi(1) = 0.8413$ to find $\displaystyle P(\text{sample mean} > 21)$.

The sample mean's spread (standard error) is σ/√n = 6/6 = 1. Standardizing 21 gives z = (21−20)/1 = 1, so P(x̄>21) = 1 − Φ(1) = 1 − 0.8413 = 0.1587. Check: 0.8413 + 0.1587 = 1.