Question

Use the appropriate normal distribution to approximate the resulting binomial distribution.

A marksman's chance of hitting a target with each of his shots is 65%. (Assume the shots are independent of each other.) If he fires 40 shots, what is the probability of his hitting the target in each of the following situations? (Round your answers to four decimal places.)

(a) at least 29 times

(b) fewer than 21 times

(c) between 23 and 29 times, inclusive

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer #1

P(hitting target), p = 0.65

q = 1 - p = 0.35

n = 40

Mean = np

= 40x0.65

= 26

Standard deviation =

=

= 3.0166

a) P(X < A) = P(Z < (A - mean)/standard deviation)

P(at least 29 times) = P(X 29)

= 1 - P(X < 28.5) *(with continuity correction)*

= 1 - P(Z < (28.5 - 26)/3.0166)

= 1 - P(Z < 0.83)

= 1 - 0.7967

= **0.2033**

b) P(fewer than 21 times) = P(X < 20.5)

= P(Z < (20.5 - 26)/3.0166)

= P(Z < -1.82)

= **0.0344**

c) P(between 23 and 29 times, inclusive) = P(22.5 < X < 29.5)

= P(X < 29.5) - P(X < 22.5)

= P(Z < (29.5 - 26)/3.0166) - P(Z < 22.5 - 26)/3.0166)

= P(Z < 1.16) - P(Z < -1.16)

= 0.8770 - 0.1230

= **0.7540**

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