Question

According to a survey in a​ country, 24% of adults do not own a credit card. Suppose a simple random sample of 700 adults is obtained.

What is the probablity that in a random sample of 700 adults between 21% and 26% do not own a credit card?

Answer 1

Let, define a random variable X that represent number of adults do not own credit card.

According to a survey in a country 24% of adults do not own a credit card.

So, the probability of a adult do not own credit card is 0.24

i.e, p=0.24

Suppose a simple random sample of 700 adults is obtained.(so, n=700)

So, the mean and variance of random variable X is defined as follow

E(X)=np=700*0.24=168

V(X)=npq=700*0.24*(1-0.24)=127.68

Here, sample is very large, so the the distribution of X is normal.

i.e, X~N(168,127.68)

Here, we need to find,What is the probablity that in a random sample of 700 adults between 21% and 26% do not own a credit card?

So, 21% of 700 adults is 147

So, 26% of 700 adults is 182

[ Z~N(0,1)]

[ from standard normal table]

[ round to 4-decimal place]

The probability is 0.8608 that in a random sample 700 adults between 21% and 26% do not own a credit card.