Question

A grocery store purchases bags of oranges from California to sell in their store. The weight of a bag of California oranges is normally distributed with a mean of 7.1 pounds and a variance of 1.21 pounds². A bag of California oranges is randomly selected in the grocery store.

(Round all probability answers to four decimal places.)

a. What is the probability that a randomly selected California orange bag purchased by a customer weighs more than 8 pounds?

b. What is the probability that a randomly selected California orange bag purchased by a customer weighs between 6.4 and 7.5 pounds?

c. What is the probability that a randomly selected California orange bag purchased by a customer weighs exactly 5 pounds?

(Round weight to two decimal places)

d. 17% of the time, a customer will buy a bag of California oranges that weighs more than a specific weight. Find that weight.

Answer 1

Solution :

Given that ,

mean = = 7.1

variance = 1.21

standard deviation = = 1.1

(a)

P(x > 8) = 1 - P(x < 8)

= 1 - P((x - ) / < (8 - 7.1) / 1.1)

= 1 - P(z < 0.81)

= 1 - 0.791

= 0.209

Probability = 0.209

(b)

P(6.4 < x < 7.5) = P((6.4 - 7.1)/ 1.1) < (x - ) / < (7.5 - 7.1) / 1.1) )

= P(-0.63 < z < 0.36)

= P(z < 0.36) - P(z < -0.63)

= 0.6406 - 0.2643

= 0.3763

Probability = 0.3763

(c)

P(x = 5 ) = 0

Probability = 0

(d)

P(Z > z) = 17%

1 - P(Z < z) = 0.17

P(Z < z) = 1 - 0.17 = 0.83

P(Z < 0.95) = 0.83

z = 0.95

Using z-score formula,

x = z * +

x = 0.95 * 1.1 + 7.1 = 8.145

Weight = 8.145