According to a survey, 67% of murders committed last year were cleared by arrest or exceptional means. fifty murders committed last year are randomly selected, and the number cleared by arrest or exceptional means is recorded.
a) find the probability that exactly 41 murders were cleared.
b)the probability that between 36 and 38 of the murders, inclusive were cleared is?
c) would it be unusual if fewer than 20 of the murders were cleared? why or why not?

a) P(X=41)= 0.0086 = 0.86%

b) P(36≤X≤38)== 0.1215 = 12.15%

c) would be unusual that fewer that 20 murders were cleared because P(X≤20) = 0.000081 =0.0081%

Step-by-step explanation:

if the random variable X= number of cleared murders , then X follows a binomial distribution:

P(X=x)= n!/[(n-x)!*x!]*p^x*(1-p)^(n-x)

where P(X=x)= probability of x cleared murders, n= number of murders selected=50 , p= probability for a murder to be cleared (67%)

therefore

a) P(X=41)=50!/[(50-41)!*41!]*0.67^41*0.33^(50-41)= 0.0086

b) P(36≤X≤38)= P(X≤38) - P(X≤36) = 0.9371- 0.8156 = 0.1215

c) P(X≤20) = 0.000081

therefore would be unusual that fewer that 20 murders were cleared

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

answer: c. is your answer

step-by-step explanation: get me a black leather coat