Question

A marksman's chance of hitting a target with each of his shots is 60%. (Assume the shots are independent of each other.) If he fires 30 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 21 times (b) fewer than 13 times (c) between 14 and 21 times, inclusive

Answer 1

Solution:

We are given n=30, p=0.60, q= 1 – p = 1 – 0.60 = 0.40

We have to use normal approximation to binomial distribution.

Mean = np = 30*0.60 = 18

SD = sqrt(npq) = sqrt(30*0.60*0.40) = 2.683282

Part a

We have to find P(X≥21) = P(X>20.5) (by using continuity correction)

P(X>20.5) = 1 – P(X<20.5)

Z = (20.5 - 18) / 2.683282

Z = 0.931695

P(Z<0.931695) = P(X<20.5) = 0.824253

P(X>20.5) = 1 – P(X<20.5)

P(X>20.5) = 1 – 0.824253

P(X>20.5) = 0.175747

Required probability = 0.1757

Part b

Here, we have to find P(X<13)

P(X<13) = P(X<12.5) (by using continuity correction)

Z = (12.5 - 18) / 2.683282

Z = -2.04973

P(Z<-2.04973) = 0.020195

Required probability = 0.0202

Part c

We have to find P(14≤X≤21)

P(14≤X≤21) = P(13.5<X<21.5) (by using continuity correction)

P(13.5<X<21.5) = P(X<21.5) – P(X<13.5)

Find P(X<21.5)

Z = (21.5 - 18) / 2.683282

Z = 1.304373

P(Z<1.304373) = P(X<21.5) = 0.903947

Find P(X<13.5)

Z = (13.5 - 18) / 2.683282

Z = -1.67705

P(Z<-1.67705) = P(X<13.5) = 0.046766

P(13.5<X<21.5) = P(X<21.5) – P(X<13.5)

P(13.5<X<21.5) = 0.903947 - 0.046766

P(13.5<X<21.5) = 0.857181

Required probability = 0.8572