Question

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 22 and a mean diameter of 200200 inches. If 8383 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 less than 0.20.2 inches? Round your answer to four decimal places. Answer

Answer 1

The probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches is 0.5319.

Explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard-deviation σ and appropriately huge random-samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

`\mu_(\bar x)=\mu`

And the standard deviation of the distribution of sample mean is given by,

`\sigma_(\bar x)=(\sigma)/(√(n))`

The information provided is:

`\mu=200\\\sigma=22\\n=83`

Since n = 83 > 30, the Central Limit Theorem can be used to approximate the sampling distribution of sample mean diameter of the shafts.

Then:

Mean:
`\mu_(\bar x)=\mu=200`

Standard deviation:
`\sigma_(\bar x)=(\sigma)/(√(n))=(22)/(√(83))=2.415`

Compute the probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches as follows:

`P(\bar X-\mu_(\bar x)<0.20)=P((\bar X-\mu_(\bar x))/(\sigma_(\bar x))<(0.20)/(2.415))`

`=P(Z<0.08)\\=0.53188\\\approx0.5319`

*Use a z-table.

Thus, the probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches is 0.5319.