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
