Question

The number of bank accounts randomly opened per month by a Zanaco customers in Lusaka is represented by the probability distribution below.
X
19
20
21
22
P(X = x)
0.4
0.25
0.2
0.15
i. Find the expected number of accounts opened per month.
ii. Suppose opening an account incurs a fixed monthly costs of K100 and an additional handling cost of k5 per account. Find the expected monthly cost of opening bank accounts by Zanaco customers.

Answer 1

solution I

Expected number of accounts per month

$$\bar x = \sum x .p(x)$$

$$=(19*0.4)+(20*0.25)+(21*0.2)+(22*0.15)$$

$$=20.1$$

answer: the expected number of accounts opened in a month is 20.1 or approximately 20 since you cannot create a fraction of an account

solution II

Fixed costs $=k100$

Cost per account $=k5$

Given expected monthly accounts opened $=20.1$

Expected monthly cost of opening accounts

$$=k100+(20.1*k5)$$ $$=k200.5$$

answer: expected monthly cost of opening accounts is k 200.5 or k 200 assuming number of accounts opened in a month are rounded off to 20 since a fraction of an account cannot be created