Let x be a binomial random variable (n = 25, p = 1/5) representing the number of
ID: 3245802 • Letter: L
Question
Let x be a binomial random variable (n = 25, p = 1/5) representing the number of correct guesses on a multiple choice test if there are 25 questions and a 1/5 probability of guessing right on each question. a. Find P(X = 5) b. Find P(X 0, 1, or 2) Let X be a binomial random variable (n = 100, p = 0.75), representing the number of job applications able to pass a computer literacy test. Assume the probability each applicant will pass is 75% and there are 100 applications considered. Find mu = E(x). the mean (or expected value). Sigma^2 = Var (X) and sigma (the variance and standard deviation). Describe the interval(mu = 2 sigma, mu + 2 sigma) where we can expect to find the value of times 95% of the time.Explanation / Answer
Answer to both the questions are below:
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5.3)
n = 25, p = .2
We apply binomial therom formula of pdf :
a. P(X=5) = 25C5*(.2^5)*(1-.2)^20 = 0.1960
b. P(X=0,1, or 2) = P(X<=2) = 0.0708
5.4)
a. Mu = E(X) = np = 100*.75 = 75
b.Sigma = sqrt(npq) = sqrt( np*(1-p)) = sqrt(.75*100*.25) = 4.33, Variance = npq = 18.75
c. The interval of 75+/- 2*18.75 or 75 +/- 37 means that within 38 and 112 we are 95% confident to find our
population mean
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