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?

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

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d

i think the correct option is the second option.