According to a survey, 62% of murders committed last year were cleared by arrest

Question

According to a survey, 62% of murders committed last year were cleared by arrest or exceptional means. Fifty murder committed last year are randomly selected, and the number cleared by arrest or exceptional means is recorded. Find the probability that exactly 40 of the murders were cleared. Find the probability that between 36 and 38 of the murders, inclusive, were cleared. Would it be unusual if fewer than 19 of the murders were cleared? Why or why not? The probability that exactly 40 of the murders were cleared is. (Round to four decimal places as needed.)

Explanation / Answer

Normal approximation to binomial distribution

Solution:

We are given, p = 0.62 and n = 50

First of all we have to check two conditions for normal approximations whether np>5 and nq>5

Here, p = 0.62, so, q = 1 – p = 1 – 0.62 = 0.38, n = 50

So, np = 50*0.62 = 31 > 5

And nq = 50*0.38 = 19 > 5

So, we can use normal approximation to binomial distribution.

Mean = np = 50*0.62 = 31

Standard deviation = sqrt (npq) = sqrt (50*0.62*0.38) = 3.4322

a)      Here, we have to find P(X=40)

Z = (40 – 31) / 3.4322 = 2.6222

For exact probability in approximation we assume,

P(X=40) = P(2.6222 – 0.05 < Z < 2.6222) = P(2.5722 < Z < 2.6722)

P(X=40) = P(2.5722 < Z < 2.6722) = P(Z<2.6722) – P(Z<2.5722)

P(X=40) = 0.996232 – 0.994947 = 0.001285

Required Probability = 0.001285

b)      Here, we have to find P(36<X<38)

P(36<X<38) = P(X<38) – P(X<36)

Z = (36 – 31) / 3.4322 = 1.4568

Z = (38 – 31) / 3.4322 = 2.0395

P(Z<1.4568) = 0.927413

P(Z<2.0395) = 0.9793

P(36<X<38) = P(X<38) – P(X<36) = P(Z<2.0395) – P(Z<1.4568) = 0.9793 – 0.9274 = 0.051887

Required Probability = 0.051887

c)      Here, we have to find P(X<19)

Z = (19 – 31) / 3.4322 = -3.4963

P(X<19) = P(Z<-3.4963) = 0.0002359

This probability is unusual because it is less than the probability 0.05 and we say that the probability is unusual if it is less than 0.05.

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