In a large population, 81% of the households have cable tv. A simple random sample of 196 households is to be contacted and the

Question

3 years ago

11

sample proportion computed. What is the mean and standard deviation of the sampling distribution of the sample proportions

1 answer:

IrinaK [193] · Answer 1 · 2022-09-15T15:30:20+03:00

Answer:

0.028

Step-by-step explanation:

Given that :

Population = Large

Household with cable TV = 81%

Sample size (n) = 196

The mean of sampling distribution (μx) = p = 81% = 81/100 = 0.81

The standard deviation of sampling distribution

Standard deviation (σ) :

σ = √[(p*(1 - p)) / n]

= √(0.81 * (1 - 0.81)) / 196

= √(0.1539 / 196)

= √0.0007852

= 0.0280214

= 0.028