58,212 views
6 votes
6 votes
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.

User CuriousMind
by
2.8k points

1 Answer

24 votes
24 votes

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=(X-\mu)/(\sigma) \\ \text{then, in our case, } \\ Z=(99-93)/(6)=(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

User Walrus The Cat
by
3.2k points