Question

A survey found that 21% of Americans watch fireworks on television on July 4th. Find the mean, variance, and standard deviation of the number of individuals who watch fireworks on television on July 4th if a random sample of 1000 Americans is selected.

Answer 1

Answer:

Mean=210

Variance=165.9

Standard deviation=12.88

Step-by-step explanation:

The survey indicates the occurrence of binomial experiment because

1. There are two possibilities can be termed as success or failure. The success would be the American watching fireworks on 4th July on TV.

2. The probability of American watching fireworks on 4th July on TV remains same for each  trial.

3. The experiment is repeated for 1000 Americans.

4. Also, the trails are independent.

The mean of binomial distribution can be achieved by multiplying the number of trails to probability of success.

$mean=np$

Here, n=1000 and p=0.21.

mean=1000*0.21

mean=210

The variance of binomial distribution can be achieved as

$variance=npq$

where q=probability of failure=1-p=1-0.21=0.79

variance=1000*0.21*0.79

variance=165.9

The standard deviation of binomial distribution can be calculated by taking square root of variance

$Standard deviation=\sqrt{npq}$

$Standard deviation=\sqrt{165.9}$

standard deviation=12.88