The average homicide rate for the cities and towns in a state is 10 per 100,000 population with a standard deviation of 2. If the variable is normally distributed, what is the probability that a randomly selected town will have a homicide rate greater than 14?

Answer:t

Pr (x>14) = 0.0228

Step-by-step explanation:

Mean (u) = 10

Standard deviation (α) = 2

Number of population (n) = 100,000

Let X be a random variable which is a measure of the town to have homicide

For normal distribution,

Z = (X - u) / α

For Pr (x=14) we have

(14 - 10) / 2

= 4/2

= 2

From the normal distribution table, 2 = 0.4772

Φ (z) = 0.4772

Recall that:

If Z is positive,

Pr (x>a) = 0.5 - Φ (z)

Pr (x>14) = 0.5 - 0.4772

= 0.0228