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Suppose a batch of steel rods produced at a steel plant have a mean length of 176 millimeters, and a variance of 100 .
If 446 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 0.65 millimeters? Round your answer to four decimal places.
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Step 1 ：We are given that the mean length of the steel rods produced at a steel plant is $176$ millimeters and the variance is $100$.

Step 2 ：We are also given that we have a sample size of $446$ rods.

Step 3 ：The standard deviation of the sampling distribution of the mean (also known as the standard error) is the standard deviation of the population divided by the square root of the sample size. In this case, the standard error is $\sqrt{100}/\sqrt{446}\approx 0.4735$.

Step 4 ：We are asked to find the probability that the mean length of the sample rods would differ from the population mean by less than $0.65$ millimeters. This is equivalent to finding the probability that the sample mean is between $176-0.65=175.35$ and $176+0.65=176.65$.

Step 5 ：We can standardize these values by subtracting the population mean and dividing by the standard error to get z-scores. The z-scores are $(175.35-176)/0.4735\approx -1.3727$ and $(176.65-176)/0.4735\approx 1.3727$.

Step 6 ：We can use the standard normal distribution to find the probabilities corresponding to these z-scores. The probabilities are approximately $0.0849$ and $0.9151$ respectively.

Step 7 ：The probability that the mean length of the sample rods would differ from the population mean by less than $0.65$ millimeters is the difference between these two probabilities, which is $0.9151-0.0849=0.8302$.

Step 8 ：Final Answer: The probability that the mean length of the sample rods would differ from the population mean by less than $0.65$ millimeters is approximately $\boxed{{\displaystyle 0.8302}}$.