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# Binominal Distribution: Probability and Normal Approximation

In a certain binomial distribution, &#960; = 0.20 and n = 30. In using the normal approximation,
a. What are the mean and standard deviation of the corresponding normal distribution?
b. If x = the number of "success" among the 30 observation, determine the following: P(x = 5), P(4 &#8804; x &#8804; 7), P(1 &#8804; x &#8804; 5), P(x &#8805; 7).

A study by the National Golf Foundation reports that the 6.2 million golfers over the age of 50 spent an average of \$939 on golf during the previous year. Assuming a normal distribution with a standard deviation of \$200m what is the probability that a randomly selected golfer in this age group will have spent:
a. more than \$1539?
b. between \$939 and \$1339?
c. less than \$1139?
d. between \$539 and \$1139?

#### Solution Preview

(1) p = 0.20, n = 30, q = 1 - p = 0.80

m = np = 30 * 0.2 = 6, s = sqrt(npq) = sqrt(30 * 0.2 * 0.8) = 2.191

(a) P(x = 5) can be found only from binomial distribution as P(5, 30) = 30C5 0.2^5 ...

#### Solution Summary

Complete, Neat and Step-by-step Solutions are provided.

\$2.19