SOLUTION: 1.Suppose we are flipping a fair coin (i.e., probability of heads = .5 and probability of tails = .5). What is the probability that in a sample of 5 flips, fewer than 4 will be hea

Algebra ->  Probability-and-statistics -> SOLUTION: 1.Suppose we are flipping a fair coin (i.e., probability of heads = .5 and probability of tails = .5). What is the probability that in a sample of 5 flips, fewer than 4 will be hea      Log On


   

Question 1020389: 1.Suppose we are flipping a fair coin (i.e., probability of heads = .5 and probability of tails = .5). What is the probability that in a sample of 5 flips, fewer than 4 will be heads?
2.The U.S. mint, which produces billions of coins annually, has a mean daily defect rate of 4 coins. Let X be the number of defective coins produced on a given day.
What is the variance of this distribution?

3. Consider the following discrete distribution. Note that x represents the value of a particular outcome and P(x) represents the probability of that outcome.
x P(x)
0 0.1
1 0.2
2 0.3
3 0.2
4 0.1
5 0.1

What is the probability that an observed x is less than or equal to 3?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
1. The statistical experiment follows the binomial distribution, with pmf p%28x%29+=+C%285%2Cx%29%2A0.5%5E%285-x%290.5%5Ex+=+C%285%2Cx%290.5%5E5. The answer is p(0) + p(1) + p(2) + p(3). But an easier way is to evaluate 1-p(4)-p(5) = 1-%28C%285%2C4%29%2BC%285%2C5%29%29%2A0.5%5E5+=+6%2A0.5%5E5 = 1-3/16 = 13/16 = 0.8125.
3. The answer is p(0) + p(1) + p(2) + p(3), but it is easier to compute 1- p(4)-p(5) = 1-0.1 - 0.1 = 0.8.