# Suppose a batch of steel rods produced at a steel plant have a mean length of 164 millimeters, and a variance of 121.?

If 287 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by less than 0.56 millimeters? Round your answer to four decimal places.

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- 4 weeks ago
variance is 121 mm? Then the standard deviation is 11 mm. Find the z-score

-0.56/11 and 0.56/11

z = -0.05090909090909090909090909090909...

z = 0.05090909090909090909090909090909....

z = -0.051

z = 0.051

48.01% of the rods will have a length less than 164 - 0.56 and 51.99% will have a length less than 164 + 0.56, which means that 51.99% - 48.01% will have a length between 164 - 0.56 and 164 + 0.56

51.99 - 48.01 = 3.98

3.98% => 0.0398

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