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

Answer:

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:

$\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373$

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.

Step-by-step 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:

$nCx=\frac{n!}{(n-x)! x!}$

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:

$\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373$

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.