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 201 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

ANSWER:

Let X denotes the number of disable persons in the sample of 1,000 persons selected at random. Then X is a binomial random variable with parameters n=1,000 and p = 0.193 as the 19.3 percent of the majority of civilians aged five and over is having some type of long-lasting condition or disability living the united states in 2000. The probability distribution function of X is given as below.

P (X = x ) = 10000 C_x (0.193)^x (1-0.193)^1000-x For x= 0,1,2,...........,,,10000.

As the sample size is large , it is complicated to find the probabilities using the binomial distribution. Use the normal approximation to find the probabilities. According to the central limit theorem, if X Bin (n,p) then for large n, X N ( np , np (1-p )). Here use the normal approximation to find the probabilities related to the persons of disability.

Using the normal approximation , the distribution of X is redefined as below.

X Bin (1000, 0, 193)

X N (1000 (0.193) , 1000 (0.193 ) ( 1-0.193 ))

X N (193,12,48² )

(a)

The probability that more than 200 persons on the sample have a disability is denoted as P(X>200). Since X is a discrete random variable and it is approximated using continuous distribution use continuity correction to find the probability. Since X is normally distributed , use the Z- transformation to find the probabilities.

( Using the complementary property)

(From Z-table.)

The probability that more than 200 persons in the sample have a disability is approximately = 0.274.

(b) The probability that between 180 persons and 300 persons in the sample have a disability is denoted as p(180<X<300). Use continuity correction and standard normal transformation to find this probability as below.

The probability that between 180 persons and 300 persons in the sample have a disability is approximately 0.860