Answer:
(a) We would expect 148.2 murders to be committed with a firearm.
(b) Yes, because 167 is greater than μ + 2σ .
Explanation:
Let X = number of murders that are committed with a firearm.
The probability that a murder is committed with a firearm is, p = 0.741.
(a)
A random sample of n = 200 murders are selected.
A murder being committed with a firearm is independent o the others.
The random variable X follows a Binomial distribution with parameters n = 200 and p = 0.741.
The expected value of a Binomial random variable is:
E (X) = n × p
Compute the expected number of murder committed with a firearm in the sample of 200 murders as follows:
E (X) = n × p
= 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 X is unusual then:
X < μ - 2σ or X > μ + 2σ
The value that is considered unusual here is,
X = 167.
Check whether 167 murders with firearm are unusual or not as follows:
μ ± 2σ = np ± (2 × √np(1- p))
= 148.2 ± 6.1955
= (142.0045, 154.3955)
≈ (142, 154)
The value 167 lies outside this range or X > μ + 2σ ⇒ 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 μ + 2σ .