Question

The diameters of ball bearings are distributed normally. The mean diameter is 7373 millimeters and the variance is 44. Find the probability that the diameter of a selected bearing is less than 7676 millimeters. Round your answer to four decimal places.

Answer 1

Answer:

0.9332

Step-by-step explanation:

We are given that

Mean diameter, $\mu=73$

Variance, $\sigma^2=4$

We have to find the probability that the diameter of a selected bearing is less than 76.

Standard deviation, $\sigma=\sqrt{variance}=\sqrt{4}=2$

$P(x<76)=P(\frac{x-\mu}{\sigma}<\frac{76-73}{2})$

$P(x<76)=P(Z<\frac{3}{2})$

Where $Z=\frac{x-\mu}{\sigma}$

$P(x<76)=P(Z<1.5)$

$P(x<76)=0.9332$

Hence, the probability that the diameter of a selected bearing is less than 76=0.9332