Suppose 40% of the population of pre-teens have a TV in their bedroom. If a random sample of 500 pre-teens is drawn from the population, then the probability that between 36% and 44% of the pre-teens have a TV in their bedroom is: a) 0.9644
b) 0.4644
c) 0.0356
d) 0.9328
e) 0.0712

0.9328

Step-by-step explanation:

The mean of the sampling distribution is equal to

up = p = 0.40

The standard deviation = √pq/n

= √0.40x0.60/500

= √0.00048

= 0.0219

We compute the z values for the probability of 36% and 44%

When p = 0.36

z = 0.36-0.40/0.0219

= -1.83

When p = 0.44

z = 0.44-0.49/0.0219

= 1.813

From the standard normal distribution table. The probability associated with z values from Which we subtract the probability

P(0.36<=p<=0.44)

= P(-1.83<=z<=1.83)

P(z<=1.83)-p(z<=-1.83)

= 0.9664-0.0336

= 0.9328

Therefore the answer is 0.9328

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