Question

According to a​ report, 70.1​% of murders are committed with a firearm. ​(a) If 400 murders are randomly​ selected, how many would we expect to be committed with a​ firearm? ​(b) Would it be unusual to observe 300 murders by firearm in a random sample of 400 ​murders? Why? ​(a) We would expect 280.4 to be committed with a firearm. ​(b) Choose the correct answer below. A. ​No, because 300 is greater than mu plus 2 sigma. B. ​No, because 300 is less than mu minus 2 sigma. C. ​Yes, because 300 is between mu minus 2 sigma and mu plus 2 sigma. D. No​, because 300 is between mu minus 2 sigma and mu plus 2 sigma . E. Yes​, because 300 is greater than mu plus 2 sigma .

Answer 1

Answer:

a) 280.4; b) E. Yes​, because 300 is greater than mu plus 2 sigma .

Step-by-step explanation:

For part a, we multiply the percentage of murders committed by a firearm by the total number of murders:

0.701(400) = 280.4

This will be the average number of murders by a firearm out of a sample of 400.

Since this is a binomial distribution (two outcomes; independent trials; same probability of success for each trial; and fixed number of trials), the standard deviation is given by

√npq

Since n = 0.701, q = 1-0.701 = 0.299.

This gives us

√(400*0.701*0.299) = √83.8396 = 9.16

This means any unusual values will be less than 2 standard deviations below the mean or more than 2 standard deviations above the mean.

2 standard deviations below the mean will be 280.4-2(9.16) = 280.4-18.32 = 262.08. 300 is not below this value.

2 standard deviations above the mean will be 280.4+298.72.  300 is above this, so it is an unusual value.