Question

1. The sales associates at TV World earn a commission on every extended warranty that they sell to customers who buy a TV, so they have an incentive to aggressively promote that warranty. Suppose that TV World sells an extended warranty to 12% of the customers who buy a TV. We are planning to randomly select 135 customers who have bought a TV from TV World in the last year.

   1. What is the sampling distribution of the sample proportion of the customers who bought an extended warranty when they bought a TV? (You should include the mean and standard deviation)

   2. What is the probability that between 8% and 13% of the customers in the sample bought an extended warranty?

   3. What is the probability that at most 20% of the customers in the sample bought an extended warranty?

   4. What is the probability that at least 20% of the customers in the sample bought an extended warranty?

Answer 1

A)

Here, μ = 0.12, σ = 0.028, x1 = 0.08 and x2 = 0.13. We need to compute P(0.08<= X <= 0.13). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.08 - 0.12)/0.028 = -1.43
z2 = (0.13 - 0.12)/0.028 = 0.36

Therefore, we get
P(0.08 <= X <= 0.13) = P((0.13 - 0.12)/0.028) <= z <= (0.13 - 0.12)/0.028)
= P(-1.43 <= z <= 0.36) = P(z <= 0.36) - P(z <= -1.43)
= 0.6406 - 0.0764
= 0.5642

b)
Here, μ = 0.12, σ = 0.028 and x = 0.2. We need to compute P(X <= 0.2). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.2 - 0.12)/0.028 = 2.86

Therefore,
P(X <= 0.2) = P(z <= (0.2 - 0.12)/0.028)
= P(z <= 2.86)
= 0.9979

c)
Here, μ = 0.12, σ = 0.028 and x = 0.2. We need to compute P(X >= 0.2). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.2 - 0.12)/0.028 = 2.86

Therefore,
P(X >= 0.2) = P(z <= (0.2 - 0.12)/0.028)
= P(z >= 2.86)
= 1 - 0.9979 = 0.0021