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

To determine the probability that the average diameter of a sampled ball bearing is greater than 0.530 cm, use the Central Limit Theorem. Calculate the standard error of the mean using the formula: SE = standard deviation / sqrt(n), then calculate the z-score using the formula: z = (sample mean - population mean) / SE. Finally, use the z-score to find the probability using a standard normal distribution table or calculator.

Step-by-step explanation:

To determine the probability that the average diameter of a sampled ball bearing is greater than 0.530 cm, we can use the Central Limit Theorem. First, we need to calculate the standard error of the mean. The formula for the standard error of the mean is:

SE = standard deviation / sqrt(n)

where SE is the standard error of the mean, standard deviation is the population standard deviation, and n is the sample size.

Given that the standard deviation is 0.008 cm and the sample size is 18, we can plug in these values to calculate the standard error:

SE = 0.008 / sqrt(18) = 0.00188 cm

Next, we need to calculate the z-score, which represents the number of standard deviations the sample mean is away from the population mean. The formula for the z-score is:

z = (sample mean - population mean) / SE

In this case, the sample mean is 0.530 cm, the population mean is 0.527 cm, and the standard error is 0.00188 cm. Plugging in these values, we can calculate the z-score:

z = (0.530 - 0.527) / 0.00188 ≈ 1.596

Finally, we can use the z-score to find the probability using a standard normal distribution table or a calculator. The probability that the average diameter of a sampled ball bearing is greater than 0.530 cm is the area under the curve to the right of the z-score.

Answer 2

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(x-\mu)/((\sigma)/(√(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(0.530-0.527)/((0.008)/(√(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.