Question

In a souvenir, 100 tourist enter each day. Assume that each tourist's decision to purchase a souvenir is independent of one another. The probability that a tourist purchases a souvenir is 40%. The probability that a tourist purchases a souvenir is 40%. The probability that a tourist purchases more than one souvenir is 0%.

(a) What is the expected number of purchases each day?

(b) What is the probability that 48 tourist make purchases on a particular day?

(c) Assume that each souvenir is sold for $8. 90% of the time, what is the minimum revenue of the shop?

(d) What is the probability that the 5th tourist of a particular day makes the first purchase of that day?

(e) On another day, 70 tourist make purchases in the souvenir shop of all the tourist that visited, 8 different tourist are surveyed. What is the probability that 5 of the surveyed. What is the probability that 5 of the surveyed tourist need a purchase?

Answer 1

(a) Probability that a tourist purchases a souvenir = 0.4

This can be treated as a binomial random variable, with n = 100, p=0.4, q = 1-0.4 = 0.6

Expected number of purchases = n*p = 100*0.4 = 40

(b) P(X=48) = C[100,48] * 0.4⁴⁸ * 0.6⁵² = 0.0215

(c) Here we need to find a 90% confidence interval for the mean.

We can approximate the binomial distribution with a normal distribution having mean, n*p = 100*0.4 = 40

and variance

To calculate a 90% confidence interval,

The 90% confidence interval is given as

Minimum revenue = 0.3194*100*8= $255.52

(d) Probability that the 5th tourist makes the first purchase = 0.6*0.6*0.6*0.6*0.4 = 0.0518

(e) On this specific day, 70 of the 100 tourists made the purchase.

Probability of a purchase on this day = 70/100 = 0.7

8 tourists are surveyed.

Probability that 5 of them did the purchase = C[8,5] * 0.7⁵ * 0.3³ = 56*0.1681*0.027 = 0.2541