According to a report, % of murders are committed with a firearm. (a) If murders are randomly selected, how many would we expect to be committed with a firearm? (b) Would it be unusual to observe murders by firearm in a random sample of murders? Why?

Full question attached

A. 230.4

B. 3.23

Explanation:

1. if 78.26% of murders are committed with firearms, 300 murders selected at random would be expected to have been committed with 300*0.7826= 230.4 firearms

2. To find how unusual this figure is, we find standard deviation from the mean 230.4

Standard deviation = x-u/N

=254-230.4/7.311= 3.23 rounded

Therefore 254 is unusual

7.311 is calculated √np(1-p)

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )