According to a survey, 62% of murders committed last year were cleared by arrest or exceptional means. Fifty murde committed last year are randomly selected, and the number cleared by arrest or exceptional means is recorded. When technology is used, use the Tech Help button for further assistance.

a. Find the probability that exactly 41 of the murders were cleared.
b. Find the probability that between 36 and 38 of the murders, inclusive, were cleared.
c. Would it be unusual if fewer than 19 of the murders were cleared? Why or why not?

a.The probability that exactly 41 of the murders were cleared is ____. (Round to four decimal places as needed.)

Answer :

Answer:

a) The probability that exactly 41 of the murders were cleared is 0.0013

b) The probability that between 36 and 38 of the murders, inclusive, were cleared is 0.0809.

c) Yes, it would be unusual

Step-by-step explanation:

Let p=62% considered as the probability of having a commited that is cleared by arres or exceptional means. We assume that choosing each of the 50 commited is independent of each choose. Then, let X be the number of cleared. In this case, X is distributed as a binomial random variable. Recall that, in this case,

[tex] P(X=k) = \binom{50}{k} p^{k}(1-p)^{50-k}[/tex] for[tex]0\leq k \leq 50[/tex], with p=0.62

a) We have that

[tex] P(X=40) = \binom{50}{40} p^{40}(1-p)^{50-40} =0.001273487  [/tex]

b) We are asked for the following

[tex]P(36\leq X \leq 38) = P(X=36)+P(X=37)+P(X=38) = 0.080888936

[/tex] (The specific calculation is omitted.

c) We will check for the following probability [tex]P(X\leq 19)[/tex]

[tex]P(X\leq 19 ) = \sum_{k=0}^{19} P(X=k) = 0.000499222 [/tex]

Given that the probability of this event is really close to 0, it would be unusual if less than 19 murders are cleared.

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