Question

Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is P(x<_ 92)?

A. 0.975
B. 0.84
C. 0.025
D. 0.16

Answer 1

\[\mathbb P(X\le92)=\mathbb P\left(\frac{X-98}{6}\le\frac{92-98}{6}\right)=\mathbb P(Z\le -1)\]where follows the standard normal distribution. You can use the empirical rule or a left-tailed \(z\) chart to look up the probability.

Answer 2

The work above is correct!
To make it clearer, note that the formula used is:
\[z=\frac{x-\mu}{\sigma} \]
where \[\mu\] is the mean and is the standard deviation