Question

According to a​ report, 78.3% of murders are committed with a firearm.​

(a) If 100 murders are randomly​ selected, how many would we expect to be committed with a​ firearm?​

(b) Would it be unusual to observe 88 murders by firearm in a random sample of 100 ​murders? Why?

Answer 1

P :- Probability of murders committed with a firearm = 0.783

Q = probability of murders not committed due to firearm = 1 - P = 1 - 0.783 = 0.217

n = 100

Part a) Using Binomial distribution

Mean of Binomial distribution is E(X) = n * P = 100 * 0.783 = 78.3

So, 78 murders are committe with firearm.

Part b) P ( X = 88 )

Probability Mass Function of Binomial distribution

Probability is unusual if probability is less than 0.05 i.e 5%

P ( X = 88 ) = 0.0051 < 0.05

It is unusual to observe 88 murders by firearm.