Question

The weight of a typical orange follows an approximately normal distribution with the mean 4.5oz and and a standard deviation of 1/2oz.

Jodie likes oranges better than apples. Her orange juice machine can only use oranges that fall within the middle 95% of weights.

what orange weights can Jodie's machine handle?

Answer 1

Solution:-

Given that,

mean = = 4.5

standard deviation = = 0.5

Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96 ) = 0.975

= z  ± 1.96

Using z-score formula,

x = z * +

x = -1.96 * 0.5 + 4.5

x = 3.52 oz

Using z-score formula,

x = z * +

x = 1.96 * 0.5 + 4.5

x = 5.48

The middle 95% are from 3.52 oz to 5.48 oz