Question

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 2.82.8 and a mean diameter of 210 inches. If 84 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.3 inches? Round your answer to four decimal places.

Answer 1

The mean diameter and standard deviation for the sample can be found using the formulas for mean and standard deviation of a sample. The probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.3 inches is 0.5.

Step-by-step explanation:

To find the mean diameter and standard deviation for the sample, we need to use the formulas for the mean and standard deviation of a sample. The mean of the sample is the same as the population mean, which is 210 inches. The standard deviation of the sample is given by the formula: standard deviation of the sample = standard deviation of the population / square root of the sample size. So, the standard deviation of the sample is 2.8 / square root of 84, which is 0.305 inches.

To find the probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.3 inches, we need to calculate the z-score for the difference of 0.3 inches and then find the corresponding probability using a z-table.

The z-score is calculated as: z = (sample mean - population mean) / (standard deviation of the sample). Plugging in the values, we get: z = (210 - 210) / 0.305 = 0. The probability that the mean diameter of the sample shafts would differ from the population mean by greater than 0.3 inches is the probability corresponding to a z-score of 0, which is 0.5.

Answer 2

Answer: 0.2322 or 23.22 %

Step-by-step explanation:

We have to solve and find the area out of these limits

μ + 0,3 = 210 + 0,3 ⇒ 210,3 and

μ - 0,3 = 210 - 0,3 ⇒ 209.7

z(l) = ( x - 210 ) / (2.8/√84) ⇒ z(l) = - (0.3 * 9,17)/ 2.8

z (l) = - 1.195

We need to interpole from z table

1.19 ⇒ 0.1170

1.20 ⇒ 0.1151

Δ ⇒ 0.01 ⇒ 0.0019

And between our point 1,195 and 1,19 the difference is 0.005

then 0.01 ⇒ 0.0019

0.005 ⇒ ?? (x)

we find x = 0.00095

to get the area for poin z (l) - 1.195 up to final left tail is from z table

0,1170 - 0.00095 = 0.1161

And by symmetry to the right is the same

So 0.1161 * 2 = 0.2322

We find the area out of the above indicated limits the area we were looking for. This is the probability of finding shafts over and below the population mean and 0.3 inches