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.

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

B. 0.9938

Step-by-step 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