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.

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:  

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

P(x > 0.530)  

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

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

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40