Question

The diameters of ball bearings are distributed normally. The mean diameter is 93 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 99 millimeters. Round your answer to four decimal places.

Answer 1

In a set with

$\begin{gathered} \operatorname{mean} \\ \mu=93 \\ \text{and standard deviation} \\ \sigma=6 \end{gathered}$

the z-score of a measure X=99, the z-score of a measure X is given by

$\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ \text{then, in our case, } \\ Z=\frac{99-93}{6}=\frac{6}{6}=1 \end{gathered}$

Now, from this z-score we need to look at the table and find the corresponding p-value. We can see that,

$\begin{gathered} P(z<1)=0.84134 \\ \text{then, } \\ P(z>1)=1-P(z<1)=1-0.84134=0.15866 \end{gathered}$

Therefore, the answer is

$P(x>99)=0.1587$