Question

A recently conducted survey found that 32% of customers stated that they prefer to buy items that they have seen advertised on television. Suppose two hundred (200) customers were randomly selected, use an appropriate approximation to determine the probability that at most 50 of the customers prefer to buy items that they have seen advertised on television.

Answer 1

Let X be the number of customers who stated that they prefer to buy items that they have seen advertised on television

The probability that a customer prefers to buy items that he/she has seen advertised on television = 0.32

X follows binomial distribution with parameters n = 200 and p = 0.32

Since n is large, we approximate X with normal distribution

Mean of X = np = 64

The standard deviation of X =

Vnp(1-p) = 6.5970

P(X < 50,5) = PX-H 50.5 – 64 o 6.5970
   We have used continuity correction

P(X < 50.5) = P(Z < -2.05)
  where Z follows a standard normal distribution

From Z table

P(Z < -2.05) = 0.0202

Therefore the probability that at most 50 of the customers out of 200 customers prefer to buy items that they have seen advertised on television is 0.0202