Question

A specific container is designed to hold 55 gallons. Suppose the actual capactiy X of a randomly chosen container of this type is normally distributed with mean 55 gallons and standard deviation 1.2 gallons.

a.) What is the probability that a randomly selected container will hold between 53.5 and 56.0 gallons?

b.) For what container capacity do 94% of containers have that capacity or larger?

c.) What is the probability that the average amount of 34 randomly selected containers is greater than 55.4 gallons?

Answer 1

Sol)

Mean = 55

S.D = 1.2

a)

Probability that container will hold between 53.5 and 56.0

Z = ( X - mean ) / S.D

= ( 53.5 - 55 ) / 1.2

= -1.25

Z = ( X - mean ) / S.D

= ( 56 - 55) / 1.2

= 0.83333

P( 53.5 < X < 56 ) = P( -1.25 < Z < 0.83333)

= P( Z < 0.8333333 ) - P( Z < -1.25 )

= 0.7997 - 0.1056

= 0.6921

b)

P( X > n ) = 94%

P( Z > ( X - mean ) / S.D ) = 0.94

( X - 55 ) / 1.2 = -1.554774

From p value to z score calculator

X - 55 = -1.8657288

X = 53.1342712

c)

34 randomly selected containers is greater than 55.4

n = 34

Z = ( X - MEAN ) / ( S.D / ✓ N)

= ( 55.4 - 55 ) / ( 1.2 /✓34)

= 0.4 / 0.20579830

= 1.943650

P( X > 55.4 ) = P( Z > 1.943650)

= 0.0247