Question

Suppose that the weight of navel oranges is normally distributed with a mean of 8 ounces and a standard deviation of 1.5 ounces.

a) What percent of navel oranges weigh between 7 ounces and 10 ounces?

b) What is the weight of the navel orange larger than only 10% of navel oranges?

Answer 1

Solution :

Given that ,

mean = = 8

standard deviation = = 1.5

(a)

P(7 < x < 10) = P((7 - 8) / 1.5) < (x - ) / < (10 - 8) / 1.5) )

= P(-0.67 < z < 1.33)

= P(z < 1.33) - P(z < -0.67)

= 0.9082 - 0.2514

= 0.6568

Answer = 65.68%

(b)

P(Z > z) = 10%

1 - P(Z < z) = 0.10

P(Z < z) = 1 - 0.10

P(Z < 1.28) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 1.5 + 8 = 9.92

weight = 9.92 ounces