Question

Consumer Reports indicated that the average life of a refrigerator before replacement is 14 years with a standard deviation of 2.5 years. Assume the age a refrigerator is replaced is normally distributed.
A.) State the random variable
B.) What is the probability that someone will keep a refrigerator fewer than 11 years before replacement?
C.) What is the probability that someone will keep a refrigerator more than 18 years before replacement?
D.) What is the probability that someone will replace a refrigerator between 8 and 15 years?
E.) Suppose a company guarantees refrigerators and will replace a refrigerator that breaks while under guarantee with a new one. However, the company does not want to replace more than 5% of the refrigerators under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)?
F.) If it turns out that 40$ of all new refrigerators need to be replaced withing 10.5 years, is there an issue with the manufacturing process? Have some numbers to back up your reasoning.

Answer 1

Answer:

a.) Average life of a refrigerator

b.) 0.1151

c.) 0.0548

d.) 0.6472225

Step-by-step explanation:

Given that :

Average life before replacement, mean, m = 14 years

Standard deviation, σ = 2 years

A.)

The random variable is the variable which is being measured.

B.) What is the probability that someone will keep a refrigerator fewer than 11 years before replacement?

Fewer than 11 years, P(x < 11)

We need to obtain the Z probability, at P(Z < 11)

The Zscore = (x - mean) / standard deviation

Zscore = (11 - 14) / 2.5 = - 1.2

P(Z < - 1.2) = 0.1151

C.) What is the probability that someone will keep a refrigerator more than 18 years before replacement?

More than 18 years, P(x > 18)

We need to obtain the Z probability, at P(Z > 18)

The Zscore = (x - mean) / standard deviation

Zscore = (18 - 14) / 2.5 = 1.6

P(Z > 1.6) = 0.0548

D.) What is the probability that someone will replace a refrigerator between 8 and 15 years?

P(x < 8)

The Zscore = (x - mean) / standard deviation

Zscore = (8 - 14) / 2.5 = - 2.4

P(Z < - 2.4) = 0.0081975

P(x < 15)

The Zscore = (x - mean) / standard deviation

Zscore = (15 - 14) / 2.5 = 0.4

P(Z < 0.4) = 0.65542

P(Z < 0.4) - P(Z < - 2.4)

0.65542 - 0.0081975 = 0.6472225