According to a report, 74.1 % of murders are committed with a firearm. (a) If 200 murders are randomly selected, how many would we expect to be committed with a firearm? (b) Would it be unusual to observe 167 murders by firearm in a random sample of 200 murders? Why? (a) We would expect nothing to be committed with a firearm. (b) Choose the correct answer below. A. No, because 167 is less than mu minus 2 sigma. B. Yes, because 167 is between mu minus 2 sigma and mu plus 2 sigma. C. Yes , because 167 is greater than mu plus 2 sigma. D. No , because 167 is between mu minus 2 sigma and mu plus 2 sigma. E. No, because 167 is greater than mu plus 2 sigma.

Question

Mathematics

Danny Horn

16 May, 13:28

According to a report, 74.1 % of murders are committed with a firearm. (a) If 200 murders are randomly selected, how many would we expect to be committed with a firearm? (b) Would it be unusual to observe 167 murders by firearm in a random sample of 200 murders? Why? (a) We would expect nothing to be committed with a firearm. (b) Choose the correct answer below. A. No, because 167 is less than mu minus 2 sigma. B. Yes, because 167 is between mu minus 2 sigma and mu plus 2 sigma. C. Yes , because 167 is greater than mu plus 2 sigma. D. No , because 167 is between mu minus 2 sigma and mu plus 2 sigma. E. No, because 167 is greater than mu plus 2 sigma.

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Answers (1)

Know the Answer?

Maliyah Baker · Answer 1

(a) We would expect 148.2 murders to be committed with a firearm.

(b) Yes , because 167 is greater than μ + 2σ.

Step-by-step 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.