Question

Part 1-

A binomial distribution has mean or expected value μ, where μ = np. What will be the expected value of number of left-handed people in a sample of 200, if the probability of one of them being left-handed is 22%?

Part 2-

A binomial distribution has a standard deviation of σ, where where σ = npq and where n is sample size, p is probability of success and q is the probability of failure. (Whatever is asked about is considered success even if it is something bad.) What is the standard deviation for a binomial distribution whose sample size n is 90, and the probability of success is 40%?

Answer #1

1)

Given

n = 200

The probability of being left-handed, p = 22% = 0.22

Now

Expected value,μ = np = 200 * 0.22 = 44

Therefore, the expected value of number of left-handed people in
a sample of 200 is **44**

**2)**

Given

n = 90

Probability of success, p = 40% = 0.40

So, probability of failure, q = 1 - p = 1- 0.40 = 0.60

Formula for standard deviation is

So,

Therefore, the standard deviation for a binomial distribution is 4.65

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