Question

A restaurant will be opening in Greenhills Shopping Center. In the process of preparing,
they have to buy appliances. A refrigerator, which costs 45,789.00, has a 15% chance of
replacement in the first three years of purchase. The store is offering an extension of the
warranty cost of 8795.00 on the refrigerator.
a. What is the expected value of the extended warranty assuming it is replaced after
three years?
b. Will you buy the refrigerator? Why?
c. Identify the mean, variance, and standard deviation.

Answer 1

a.

X=800; P(x)=0.2

X'=112.10; P(x’)=(1−0.2)

xP(x)=800∗0.2=160

x'P(x')=112.10∗0.8=89.68

Expected value $= 687.90 *0.20 +(- 112.10) *0.80 =\$47.90$

b.

So in terms of the refrigerator, it makes sense to purchase the extended warranty.

c.

The mean of a binomial distribution is given by $n * p$

Where n = 15 and p = 0.1

Hence the mean = 15 * 0.1

= 1.5

Variance

The variance of a binomial distribution is given by $n * p * q$

Where n =15, p = 0.1 and q = 1 – p = 1 – 0.1 = 0.9

Hence the variance = $15 * 0.1 * 0.9$

= 1.35

Standard deviation

Standard deviation is given by the square root of the variance,

$SD = \sqrt{(1.35)}\\ = 1.161895004\\ =1.1619$