Question

As part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket. The probability that a shopper will buy a packet of crackers after tasting the free sample is 0.200. Different shoppers can be regarded as independent trials. If X is the number among the next 100 shoppers who buy a packet of the crackers after tasting a free sample, then the probability that fewer than 30 buy a packet after tasting a free sample is approximately (Use Normal approximation to solve the problem, if its conditions are met.)

A. 0.2000
B. 0.9938
C. None of the answers are correct.

Answer 1

The probability that fewer than 30 buy a packet after tasting a free sample is approximately

B. 0.9938

Explanation:

We have

p = 0.2

q = 1 - 0.2 = 0.8

n = 100

and we need to compute P(X<30). First we need to check if normal distribution can be used or not. If np>5 then we can use normal distribution to solve this problem.

np = 100*0.2 = 20 > 5.

So we can use the normal approximation to solve this problem.

μ = np = 20

σ = √(npq)

= √(100)(0.2)(0.8)

σ = 4

We know that z = (X - μ)/σ, So

P(X<30) = P[(X-μ)/σ < (30 - μ)/σ]

= P(z<(30-20)/4)

= P(z < 10/4)

= P (z<2.5)

Using the normal distribution probability table, we get:

P(z<2.5) = 0.9938

So, the correct option is B. 0.9938