Question

In the United States, 35% of households own a 4K television. Suppose we take a random sample of 150 households. (a) Describe the distribution of the sample proportion. (b) What is the probability that in this sample of 150 households that more than 50% own a 4K television

Answer 1

(a) The distribution of the sample proportion is Normal distribution.

(b) The probability that in this sample of 150 households that more than 50% own a 4K television is 0.00012.

Explanation:

We are given that in the United States, 35% of households own a 4K television.

Suppose we take a random sample of 150 households.

Let
`\hat p` = sample proportion of households who own a 4K television.

The z-score probability distribution for sample proportion is given by;

Z =
`\frac{ \hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion

p = population proportion of households own a 4K television = 35%

n = sample of households = 150

(a) The distribution of the sample proportion is related to the Normal distribution.

(b) Probability that in this sample of 150 households more than 50% own a 4K television is given by = P(
`\hat p` > 0.50)

P(
`\hat p` > 0.50) = P(
`\frac{ \hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` >
`\frac{ 0.50-0.35}{\sqrt{(0.50(1-0.50))/(150) } }` ) = P(Z > 3.67) = 1 - P(Z
`\leq` 3.67)

= 1 - 0.99988 = 0.00012

Now, in the z table the P(Z
`\leq` x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 3.67 in the z table which has an area of 0.99988.

Therefore, probability that in this sample of 150 households more than 50% own a 4K television is 0.00012.