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 0.008 cm. Suppose that a sample of 18 ball bearings are randomly selected from a very large lot. Determine the probability that the average diameter of a sampled ball bearing is greater than 0.530 cm.

Answer 1

Answer:

0.056 is the probability that average diameter of a sampled ball bearing is greater than 0.530 cm.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0.527 cm

Standard Deviation, σ = 0.008 cm

Sample size, n = 18

We are given that the distribution of diameter of a ball bearing is a bell shaped distribution that is a normal distribution.

Formula:  

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$  

P(average diameter of a sampled ball bearing is greater than 0.530 cm)  

P(x > 0.530)  

$P( x > 0.530) = P( z > \displaystyle\frac{0.530-0.527}{\frac{0.008}{\sqrt{18}}}) = P(z > 1.5909)$  

Calculation the value from standard normal z table, we have,  

$1 - P(z < 1.5909) =1 - 0.944 =0.056= 5.6\%$

Thus, 0.056 is the probability that average diameter of a sampled ball bearing is greater than 0.530 cm.