Answer:
(a) We would expect 148.2 murders to be committed with a firearm.
(b) Yes, because 167 is greater than <em>μ </em>+<em> </em>2<em>σ</em><em> </em>.
Step-by-step explanation:
Let <em>X</em> = number of murders that are committed with a firearm.
The probability that a murder is committed with a firearm is, <em>p</em> = 0.741.
(a)
A random sample of <em>n</em> = 200 murders are selected.
A murder being committed with a firearm is independent o the others.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 200 and <em>p</em> = 0.741.
The expected value of a Binomial random variable is:
E (<em>X</em>) = <em>n</em> × <em>p</em>
Compute the expected number of murder committed with a firearm in the sample of 200 murders as follows:
E (<em>X</em>) = <em>n</em> × <em>p</em>
<em> </em>= 200 × 0.741
= 148.2
Thus, the expected number of murder committed with a firearm is 148.2.
(b)
According to the rule of thumb, data values that are more than two standard deviations away from the mean are considered as unusual.
That is, if <em>X</em> is unusual then:
<em>X </em><<em> μ </em>-<em> 2σ</em> or <em>X </em>><em> μ </em>+<em> 2σ</em>
The value that is considered unusual here is,
<em>X</em> = 167.
Check whether 167 murders with firearm are unusual or not as follows:
<em>μ </em>±<em> 2σ</em> = np ± (2 × √np(1- p))
= 148.2 ± 6.1955
= (142.0045, 154.3955)
≈ (142, 154)
The value 167 lies outside this range or <em>X </em>><em> μ </em>+<em> 2σ </em>⇒ 167 > 154.
Thus, concluding that it would be unusual to observe 167 murders by firearm in a random sample of 200 murders.
Correct option:
Yes, because 167 is greater than <em>μ </em>+<em> </em>2<em>σ</em><em> </em>.