Question

(NORMAL APPOXIMATION TO BINOMIAL)

A brokerage survey reports that 27% of all individual investors have used a discount broker (one that does not charge the full commission). If a random sample of 70 individual investors is taken, approximate the probability that fewer than 21 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

Solution:

Given that,

P = 0.27

1 - P = 0.73

n = 70

Here, BIN ( n , P ) that is , BIN (70 , 0.27)

then,

n*p = 70 * 0.27 = 18.9 > 5

n(1- P) = 70 * 0.73 = 51.1 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 18.9

Standard deviation = =n*p*(1-p) = 70 * 0.27 * 0.73 = 13.797

We using continuity correction factor

P(X < a ) = P(X < a - 0.5)

P(x < 20.5) = P((x - ) / < (20.5 - 18.9) / 13.797)

= P(z < 0.43)

Probability = 0.666