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?

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lucianoangelini92 lucianoangelini92 · Answer 1 · 2019-12-03T15:45:01+01:00

Answer:

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