Question

A survey of adults ages 18-29 found that 86% use the Internet. You randomly select 125 adults ages 18-29 and ask them if they use the Internet. (a) Find the probability that exactly 102people say they use the Internet. (b) Find the probability that at least 102people say they use the Internet. (c) Find the probability that fewer than 102people say they use the Internet. (d) Are any of the probabilities in parts (a)-(c) unusual? Explain.

Answer 1

Answer:

Given that:

A survey of adults ages 18-29 found that 86% use the Internet. You randomly select 125 adults ages 18-29 and ask them if they use the Internet

Let Y be the number of people say they use the internet.

Random variable Y follows binomial distribution with parameters.

n = 125

p = 0.86

consider

np = 125*0.86 = 107.5

n(1-p) = 125(1-0.86)

= 125*0.14

= 17.5

Here, Both np >5 and n(1-p) > 5 , so the binomial random variable

Y is approximately normally distributed with

= 107.5

(a) Probability that exactly 102 people say they use the Internet.

P(X = 102) = P(101.5Y102.5)

= P(101.5-107.5/3.8794Y-/102.5-107.5/3.8794)

= P(-1.54-1.28)

= 0.0385

(b)  probability that at least 102 people say they use the Internet

P(X 102) = P(Y101.5)

= P(Y-/101.5-107.5/3.8794)

= P(Z-1.54)

= P(Z1.54)

= 0.9382

(c) probability that fewer than 102 people say they use the Internet.

P(X <102) = P(Y<101.5)

= P(Y-/<101.5-107.5/3.8794)

= P(Z<-1.54)

= 0.0168

(d) If probability of any event is less than 0.01(or 0.05) then the event to be considered as unusual event.

Here, the obtained probabilities are not too small so none of the probabilities is unusual.