Question

14-20. The owner of a newspaper stand receives 50 copies of Al Ahram newspaer eves morning. The number of copies sold, x, varies randomly according to the folou probability distribution: 4535, 36,...,49 x 50, 51,.. . ,59 33 60, 61,...,70 p(x (a) Determine the probability that the owner will sell out completely (b) Determine the expected number of unsold copies per day (e) A single copy costs 50 cents and sells for $1.00. Unsold copies have Determine the expected net income per day.

Answer 1

(a)

Probability that the owner will sell out completely = P(X $\ge$ 50)

= $\sum_{x = 50}^{59} (1/30) + \sum_{x = 60}^{70} (1/33)$

= (10/30) + (11 / 33) = 2/3 = 0.67

(b)

Let x be the number of sold copies.

Expected value of sold copies = E(x)

$= \sum_{x = 35}^{49} (x/45) + \sum_{x = 50}^{59} (x/30) + \sum_{x = 60}^{70} (x/33)$

Sum of arithmetic progression for first term = a, number of terms = n and common difference d = 1 is,

(n/2) * [2a + (n-1)d]

$\sum_{x = 35}^{49} x = (15/2) * (2 * 35 + (15-1)* 1 ) = 630$

$\sum_{x = 50}^{59} x = (10/2) * (2 * 50 + (10-1)* 1 ) = 545$

$\sum_{x = 60}^{70} x = (11/2) * (2 * 60 + (11-1)* 1 ) = 715$

E(x) = (630/45) + (545/30) + (715/33) = 53.83

Number of unsold copies = 50 - x

Expected number of unsold copies per day = E(50 - x) = 50 - E(x) = 50 - 53.83 = -3.83

(c)

Net income per day = x * ($1 - $0.5) = 0.5x dollars

As, the expected number of sold copies is greater than 50, it is expected that all copies are sold.

Expected net income per day = 0.5 * 50 = 25 dollars.