Question

Assume that 8​% of people are​ left-handed. We select 6 people at random. ​a) How many lefties do you​ expect? ​b) With what standard​ deviation? ​c) If we keep picking people until we find a​ lefty, how long do you expect it will​ take?

Answer 1

Answer:

a) 0.48

b) 0.6645

c) 12.5

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are left-handed, or they are not. The probabilities of each person being left-handed are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

The number of trials expected to find r sucesses is given by

$N = \frac{r}{p}$

In this problem we have that:

Assume that 8​% of people are​ left-handed. We select 6 people at random. ​

This means that $p = 0.08, n = 6$

a) How many lefties do you​ expect?

$E(X) = np = 6*0.08 = 0.48$

​b) With what standard​ deviation?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{6*0.08*0.92} = 0.6645$

​c) If we keep picking people until we find a​ lefty, how long do you expect it will​ take?

Number of trials to find 1 success. So

$N = \frac{r}{p} = \frac{1}{0.08} = 12.5$