Question

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.

Use normal approximation. Round the answers to four decimal places (e.g. 98.7654).

(a) Approximate the probability that more than 206 persons in the sample have a disability.

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

Answer 1

This is a binomial distribution question with
n = 1000
p = 0.193
q = 1 - p = 0.807
This binomial distribution can be approximated as Normal distribution since
np > 5 and nq > 5
Since we know that

P(x > 206.0)=?
The z-score at x = 206.0 is,

z = 1.0417
This implies that
P(x > 206.0) = P(z > 1.0417) = 1 - 0.8512


P(180.0 < x < 300.0)=?

This implies that
P(180.0 < x < 300.0) = P(-1.0417 < z < 8.5737) = 0.8512
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