Question

question 9 of 10
suppose a normal distribution has a mean of 98 and a
standard deviation of 6. what is ( p(x leq 104) )?
a. 0.16
b. 0.84
c. 0.975
d. 0.025

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 98\) (mean), \(\sigma=6\) (standard deviation), and \(x = 104\).

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

Step2: Use the standard normal distribution table

The standard normal distribution table (or the empirical rule for a normal distribution: about 68% of the data is within \(z=\pm1\), 95% within \(z = \pm2\), 99.7% within \(z=\pm3\)).
The probability \(P(Z\leq z)\) for \(z = 1\) is \(P(Z\leq1)=0.8413\approx0.84\) (using the standard normal table or the fact that for a normal distribution \(P(X\leq\mu+\sigma)\) corresponds to the area to the left of \(z = 1\) which is approximately \(0.84\) from the properties of the normal distribution).

Answer:

B. 0.84