Question

TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household in a recent year was 2.24. Assume the standard deviation is 1.2. A sample of 85 households is drawn.

1. What is the probability that the sample mean number of TV sets is between 2.5 and 3?

2. Find the 30th percentile of the sample mean.

Answer 1

Solution :

= / n = 1.2 / 85 = 0.1302

= P[(2.5 - 2.24) / 0.1302 < ( - ) / < (3 - 2.24) / 0.1302)]

= P(2.00 < Z < 5.84)

= P(Z < 5.84) - P(Z < 2.00)

= 1 - 0.9772

= 0.0228

Probability = 0.0228

Using standard normal table,

P(Z < z) = 30%

P(Z < -0.52) = 0.3

z = -0.52

= z * +   

= -0.52 * 0.1302 + 2.24 = 2.17

30th percentile = 2.17