Question

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 1

To find the probability that a randomly selected town will have a homicide rate greater than 14, we need to standardize the value of 14 using z-scores. The z-score formula is z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. For this problem, the mean (μ) is 10 and the standard deviation (σ) is 2. Therefore, the z-score is z = (14 - 10) / 2 = 2. To find the probability of a z-score greater than 2, we can use a standard normal distribution table or a calculator. Using a standard normal distribution table, the probability is approximately 0.0228 or 2.28%.

Step-by-step explanation:

To find the probability that a randomly selected town will have a homicide rate greater than 14, we need to standardize the value of 14 using z-scores. The z-score formula is z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

For this problem, the mean (μ) is 10 and the standard deviation (σ) is 2. Therefore, the z-score is z = (14 - 10) / 2 = 2. To find the probability of a z-score greater than 2, we can use a standard normal distribution table or a calculator. Using a standard normal distribution table, the probability is approximately 0.0228 or 2.28%.

Answer 2

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