Question

A brokerage survey reports that 28% of all individual investors have used a discount broker (one that does not charge the full commission). If a random sample of 85 individual investors is taken, approximate the probability that at most 20 have used a discount broker. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.

Answer 1

Let x the number of investors have used a discount broker.

x follows binomial distribution with n = 85 and p = 0.28 , q = 1-0.28 = 0.72

As n*p = 85*0.28 = 23.8 and n*q = 85*0.72 = 61.2  both are greater than 10, so we can use normal approximation to the binomial distribution.

Mean µ = n*p = 23.8

Standard deviation σ =
n*p*q
= $\sqrt{85*0.28*0.72}$ = 4.1396

We are asked to find

P( x ≤ 20 )

So we have to use continuity correction by adding 0.5 to 20

So we have to find P( x ≤ 20.5)

= $P\left ( \frac{x-\mu }{\sigma } \leq\frac{20.5-23.8}{4.1396}\right )$

= P( z ≤ -0.8 )

= 0.212 --- ( From z score table )

The probability that at most 20 have used a discount broker is 0.212