Question

It is reported that 76% of Canadians shop online. A random sample of 500 Canadians was drawn across the country to investigate this. What is the probability that more than 72% but less than 81% of Canadians shop online in this sample? Make sure you check the required condition(s) to validate your answer.  

Answer 1

The normal distribution can be used as an approximation to the Binomial distribution under certain circumstances,

If X~ B(n,p) and if n is large and p is close to 1/2

Then X is approximately N(np, npq)

Where q= 1-p.

Given that p= 0.76 , n=500

By Normal approximation,

X ~ N(np= 380, npq = 91.2)

Transformation of X variable into standard normal variable (Z)

Z~ (X-380)/√(91.2) ~ N(0,1)

Now to find,

P(0.72< X<0.81) =P( -39.7157 < Z < -39.7063)

=P(Z> 39.7063) - P(Z> -39.7157)

=1-P( Z< 39.7063) - P( Z< 39.7157)

=1-0-0

=1