Question

2. Suppose that 10% of the general population is left-handed (a) We randomly sample 20 individuals. Find the probability that fewer than 2 of them are left handed. Hint: You cannot use the normal distribution (b) we randomly sample 200 individuals. Find the probability that less than 8% of them are left handed.

Answer 1

Let X is a random variable shows the number of left handed people. Here X has binomial distribution with parameters n=20 and p=0.10.

(a)

The probability that fewer than 2 of them are left handed is

P(X < 2) (0.10)"(1-0, 10)20-1-0.3917

(b)

Here we have

p = 0.1, n=200

The sampling distribution of sample proportion will be approximately normal with mean

up p = 0.1

and standard deviation

Vpl-P) = (1-1)-. /0:1 . 0.9 0.2009 = 0.0212 0.02 12 200-=

The z-score for is

p = 0.08

0.08 -0.1 0.0212 -0.94

The required probability is

P(p < 0.08) = P(z <-0.94) = 0.1736