Question

Based on historical data, the diameter of a ball bearing is normally distributed with a mean of 0.527 cm and a standard deviation of 0.009 cm. Suppose that a sample of 36 ball bearings are randomly selected. Determine the probability that the average diameter of a sampled ball bearing is greater than 0.530 cm.
a. 0.9772
b. 0.0228
c. 0.5062
d. 0.0559

Answer 1

Mean of sample ()= 0.527 cm

The standard deviation of the sample () = 0.009 cm

Number of observations per sample (n) = 36

Let Z be the value of the Z statistic from the standard normal distribution table.

Now let us consider an average diameter of 0.530cm.

$0.530 = \mu +Z*\frac{\sigma }{\sqrt{n}}$

$0.530 = 0.527+Z*\frac{0.009}{\sqrt{36}}$

$0.530 - 0.527=Z*\frac{0.009}{6}$

$Z = \frac{0.003 * 6}{0.009}$

$Z = 2$

The probability to the left side of the Z statistic(2) on a standard normal distribution table is 0.97725

The probability that the average diameter of a sampled ball bearing is greater than 0.530 cm = 1 - 0.97725

= 0.02275

Hence the correct answer is option B.