Question

question 10 of 25
suppose a normal distribution has a mean of 79 and a
standard deviation of 7. what is ( p(xgeq93) )?
a. 0.16
b. 0.025
c. 0.84
d. 0.975

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 93\), \(\mu=79\), and \(\sigma = 7\).

$$z=\frac{93 - 79}{7}=\frac{14}{7}=2$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\geq93)\), which is equivalent to \(P(Z\geq2)\) in the standard normal distribution.
Since the total area under the standard normal curve is \(1\), and \(P(Z < 2)\) (from the standard normal table) is \(0.9772\).

$$P(Z\geq2)=1 - P(Z < 2)=1-0.9772 = 0.0228\approx0.025$$

Answer:

B. 0.025