Question

An office furniture store is having a sale on a certain type of office chair. The chairs are priced at GH¢800 per chair. The probability distribution for the number of chairs sold to an individual customer is

Number of chairs	1	2	3	4
Probability	0.15	0.40	0.10	0.35

1. a) Find the probability that the number of chairs sold to an individual customer is odd.

2. b) Find the expected number and standard deviation of chairs sold to an individual customer.

3. c) Find the expected amount and standard deviation of the amount a customer pays for the chairs he or she buys.

4. d) The delivery charge for an order of chairs is GH¢100. Find the expected amount and standard deviation of the amount a customer pays for the chairs he or she buys, including the delivery charge.

Answer 1

a)Let X be the number of chairs sold

Probability that the number of chairs sold to an individual customer is odd

= P( X=1 or X=3)

= P( X=1) +P( X=3 )

= 0.15 + 0.10

= 0.25

b) Let X be the number of chairs sold

E(X)= xP(X=x)

= 1*0.15 +2*0.40 +3*0.10+ 4*0.35

=2.65

V(X) = x^2.P(X=x) - E²(X)

= 1*1*0.15+2*2*0.40 + 3*3*0.10 +4*4*0.35 - 2.65*2.65

= 1.2275

Standard deviation = 1.1079

Expected number of chairs sold = 2.65

Standard devaition of number of chair sold = 1.1079

c) Each chair costs = 800

Expected number of chairs bought = 2.65

Standard devaition of number of chair bought = 1.1079

Expected amount the customer  pays for the chairs = 2.65*800 = 2120

Standard deviation of amount the customer  pays for the chairs = 1.1079*800 = 886.34

Note : V(800X) = 800² *V(X)

Thus , standard deviation of 800X = 800* standard deviation of X

d) Expected cost of chairs = 2120

Delivery charge = 100

Expected amount the customer pays including delivery charge = 2120 +100 = 2220

Standard deviation of amount  the customer pays including delivery charge = 886.34

Note :Note : E(X+100) = E(X) +100

V(X+100) = V(X) +V(100) = V(X) +0 = V(X)

Variance and standard deviation remains same .