Question

An orange juice producer sources his oranges from a large grove. The amount of juice extracted from a single orange is normally distributed with mean 141 ml and standard deviation 12 ml.

(a) What is the probability that a randomly-chosen orange will contain less than 150 ml of juice?

(b) What amount of juice would only 1 in 100 oranges exceed?

(c) Describe the probability distribution of the average amount of juice from a sample of 20 oranges? What is the probability that this average is less than 150ml?

Answer 1

(a)

= 141

= 12

To find P(X<150):

Z = (150 - 141)/12

= 0.75

By Technology, Cumulative Area Under Standard Normal Curve = 0.7734

So,

the probability that a randomly-chosen orange will contain less than 150 ml of juice = 0.7734

(b)

The amount of juice would only 1 in 100 oranges exceed corresponds to area = 0.50 - 0.01 = 0.49 from mid value to Z on RHS.

Table of Area Under Standard Normal Curve gives Z = 2.33

So,

Z = 2.33 = (X - 141)/12

So,

X = 141 + (2.33 X 12) = 168.96

So,

amount of juice would only 1 in 100 oranges exceed = 168.96 ml

(c)

(i)

n = 20

SE = /

= 12/

= 2.6833

So,

the probability distribution of the average amount of juice from a sample of 20 oranges is Normal Distribution with mean 141 and standard deviation 2.6833

(ii)

To find P(<150):
Z = (150 - 141)/2.6833

= 3.3541

By Technology, Cumulative Area Under Standard Normal Curve = 0.9996

So,

the probability that this average is less than 150ml = 0.9996