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?

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:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

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

[tex]N = \frac{r}{p}[/tex]

In this problem we have that:

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

This means that [tex]p = 0.08, n = 6[/tex]

a) How many lefties do you expect?

[tex]E(X) = np = 6*0.08 = 0.48[/tex]

b) With what standard deviation?

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{6*0.08*0.92} = 0.6645[/tex]

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

[tex]N = \frac{r}{p} = \frac{1}{0.08} = 12.5[/tex]

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?

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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?