Solution :
Given :
Mean, μ = 10 inches
Standard deviation, σ = 1.6 inches
Sample size is n = 39
Therefore,
[tex]$\mu_{\overline x}=\mu = 10$[/tex]
[tex]$\sigma_{\overline x}=\frac{\sigma}{\sqrt n } = \frac{1.6}{\sqrt{39}}$[/tex]
= 0.25
[tex]$P (\overline X > 10.5 ) = P\left( \frac{\overline X - \mu_{\overline x}}{\sigma_{\overline x}} > \frac{10.5 - 10}{0.25} \right)$[/tex]
= P( Z >2)
= 1 - P(Z < 2)
= 1 - 0.97225 (from standard normal table)
= 0.0277
