To solve these problems, we can use the properties of the normal distribution and the central limit theorem.
a. To find the probability that a random sample of 16 policyholders will have a mean insurance policy cost between $1250 and $1400, you first need to standardize the values using the z-score formula:
Z = (X - μ) / (σ / √n)
Where:
X is the value you want to find the probability for,
μ is the mean of the distribution,
σ is the standard deviation of the distribution,
n is the sample size.
In this case, X1 = $1250, X2 = $1400, μ = $1370, σ = $350, and n = 16.
For X1:
Z1 = (1250 - 1370) / (350 / √16)
Z1 = (-120) / (350 / 4)
Z1 = -1.37 (rounded to two decimal places)
For X2:
Z2 = (1400 - 1370) / (350 / √16)
Z2 = (30) / (350 / 4)
Z2 = 1.03 (rounded to two decimal places)
Now, you can find the probability that the sample mean falls between these two z-scores using a standard normal distribution table or calculator:
P(-1.37 < Z < 1.03) ≈ P(Z < 1.03) - P(Z < -1.37)
You can look up these values in a standard normal distribution table or use a calculator to find the probabilities.
b. To find the probability that a random sample of 16 policyholders will have a mean insurance policy cost exceeding $1450, you need to calculate the z-score for $1450 and then find the probability of Z being greater than that value.
Z = (1450 - 1370) / (350 / √16)
Calculate the value of Z, and then find P(Z > Z-value) using a standard normal distribution table or calculator.
c. To find the probability that a random sample of 16 policyholders will have a mean insurance policy cost below $1500, you need to calculate the z-score for $1500 and then find the probability of Z being less than that value.
Z = (1500 - 1370) / (350 / √16)
Calculate the value of Z, and then find P(Z < Z-value) using a standard normal distribution table or calculator.
Keep in mind that you'll need to use a standard normal distribution table or calculator to find the actual probabilities based on the calculated z-scores.