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%?
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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