Suppose a normal distribution has a mean of 79 and a standard deviation of 7. What is P(x ≥ 93)?A.0.025B.0.16C.0.84D.0.975SUBMITarrow_backPREVIOUS

To solve this problem, we need to convert the raw score (x) to a z-score. The z-score is a measure of how many standard deviations an element is from the mean.

The formula for calculating the z-score is:

Z = (X - μ) / σ

where: X is the raw score, μ is the population mean, σ is the standard deviation.

Substituting the given values into the formula, we get:

Z = (93 - 79) / 7 = 2

Now, we need to find the probability that the z-score is greater than or equal to 2. We can look this up in a standard normal distribution table, or use a calculator that can calculate it.

The value for P(Z < 2) is approximately 0.9772. However, we want P(Z ≥ 2), so we subtract the value from 1 (since the total probability under the curve is 1):

P(Z ≥ 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228

So, the answer is approximately 0.0228, which is closest to option A.0.025.