Question

The 2018 Canadian internet use survey showed, overall, 94% of Canadians had home internet access, and about 84% of the internet users bought goods or services online, spending $57.4 billion. Suppose 64 internet users with home internet access were randomly and independently sampled. Find the probability that at least 20% of those sampled currently did not buy goods or services online. Round your answers to the nearest ten-thousandth (4 decimals).

Answer 1

We first find the probability of people who did not buy goods or services online P(In and B'). For this there are two steps, first the person should be internet user P(In) and if using the net he/she should not have bought online P(B' | In) .

Given

P(In) = 94%

internet users bought goods or services online =P(B | In) = 84%

internet users not bought goods or services online =P(B' | In) = 1 - 84% = 16%

Therefore P(I and B') = P(B'| In) * P(In) ....using the formula of conditional probability

= 16% * 94%

= 0.1504

Next we treat the event as a binomial distribution where you either buy online or you don't.

Now we can treat this simply as probability of not buying goods online.

p= 0.1504 n = 64

Since n > 30 and np > 10, we can use normal approximation

E(X) = np

= 9.6256

V(X)=np(1-p)

= 8.1779

S(X) = 2.8597

At least 20% of 64 means at least 12.8

P( X >= 12.8) = P( X > 12.8)  

= P( Z > 1.11)

= 1 - P( Z < 1.11)

= 1 - 0.86651 .....................using normal distribution tables.

At least 20% = 0.13349

We would normally use continuity correction since we are approximating a discrete distribution to a continuous ne. Since here 'x' was an integer so it was used directly.