Question

According to a​ survey, 65​% 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. 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 18 of the murders were​ cleared? Why or why​ not?

Answer 1

a)
`P(X=41)=(50C41)(0.65)^(41) (1-0.65)^(50-41)=0.00421`

b)
`P(X=36)=(50C36)(0.65)^(36) (1-0.65)^(50-36)=0.0714`

`P(X=37)=(50C37)(0.65)^(37) (1-0.65)^(50-37)=0.0502`

`P(X=38)=(50C38)(0.65)^(38) (1-0.65)^(50-38)=0.0319`

And adding these values we got:

`P(36 \leq X \leq 38)= 0.1535`

c) We can find the expected value given by:

`E(X) = np =50*0.65 = 32.5`

And the standard deviation would be:

`\sigma = √(np(1-p)) √(50*0.65*(1-0.65))= 3.373`

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

`\mu -2\sigma = 32.5- 2*3.373 = 25.75`

And then we can consider a value of 18 as unusual lower for this case.

Explanation:

Let X the random variable of interest "number cleared by arrest or exceptional", on this case we can model this variable with this distribution:

`X \sim Binom(n=50, p=0.65)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Part a

We want this probability:

`P(X=41)=(50C41)(0.65)^(41) (1-0.65)^(50-41)=0.00421`

Part b

We want this probability:

`P(36 \leq X \leq 38)`

We can find the individual probabilities:

`P(X=36)=(50C36)(0.65)^(36) (1-0.65)^(50-36)=0.0714`

`P(X=37)=(50C37)(0.65)^(37) (1-0.65)^(50-37)=0.0502`

`P(X=38)=(50C38)(0.65)^(38) (1-0.65)^(50-38)=0.0319`

And adding these values we got:

`P(36 \leq X \leq 38)= 0.1535`

Part c

We can find the expected value given by:

`E(X) = np =50*0.65 = 32.5`

And the standard deviation would be:

`\sigma = √(np(1-p)) √(50*0.65*(1-0.65))= 3.373`

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

`\mu -2\sigma = 32.5- 2*3.373 = 25.75`

And then we can consider a value of 18 as unusual lower for this case.