There were 49.7 million people with some type of long-lasting condition or disability living in the United States in 2000. This represented 19.3 percent of the majority of civilians aged five and over (http://factfinder.census.gov). A sample of 1000 persons is selected at random.

(a) Approximate the probability that more than 200 persons in the sample have a disability.

(b) Approximate the probability that between 180 and 300 people in the sample have a disability.

Step 1 of 4:

Here 19.3 percent of the majority of civilians aged five and over.

A sample of 100 persons is selected at random.

We have to find the approximate probability that more than 200 persons in the sample have a disability.We have to find the approximate probability that between 180 and 300 people in the sample have disability.

Step 2 of 4:

The probability that majority of civilians aged five and over, p= 0.193.

The total number in the sample, n=1000.

Let X be the number of civilians who aged five and over.

From the given information, it is clear that X~B(1000, 0.193)

So the mean of X,

np= (10000.193).

= 193

The variance of X,

np(1-p)= (10000.1930.807).

= 155.75

Since np>5 and n(1-p)>5, normal approximation is valid.

If X is a binomial random variable with parameters n and p,

Z=

Is approximately a standard normal variable.