Question

question 10 of 10
suppose a normal distribution has a mean of 98 and a
standard deviation of 6. what is ( p(x geq 104) )?

a. 0.16

b. 0.475

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 \(\mu = 98\), \(\sigma=6\), and \(x = 104\).

$$z=\frac{104 - 98}{6}=\frac{6}{6}=1$$

Step2: Use the standard normal distribution table

We know that for a standard normal distribution \(Z\sim N(0,1)\), \(P(Z\geq1)=1 - P(Z < 1)\).
From the standard normal distribution table, \(P(Z < 1)=0.8413\approx0.84\).
So \(P(Z\geq1)=1 - 0.84=0.16\)

Answer:

A. 0.16