Question

A manufacturer knows that their items have a lengths that are skewed right, with a mean of 10 inches, and standard deviation of 1.6 inches. If 39 items are chosen at random, what is the probability that their mean length is greater than 10.5 inches

Answer 1

Given :

Mean, μ = 10 inches

Standard deviation, σ = 1.6 inches

Sample size is n = 39

Therefore,

`$\mu_(\overline x)=\mu = 10$`

`$\sigma_(\overline x)=(\sigma)/(\sqrt n ) = (1.6)/(√(39))$`

= 0.25

`$P (\overline X > 10.5 ) = P\left( (\overline X - \mu_(\overline x))/(\sigma_(\overline x)) > (10.5 - 10)/(0.25) \right)$`

= P( Z >2)

= 1 - P(Z < 2)

= 1 - 0.97225 (from standard normal table)

= 0.0277