Mathematics, 12.03.2020 06:04 coolcat3190

Suppose a batch of steel rods produced at a steel plant have a mean length of 170 millimeters, and a standard deviation of 10 millimeters. If 299 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 0.7 millimeters

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77.38% probability that the mean length of the sample rods would differ from the population mean by less than 0.7 millimeters

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 170, \sigma = 10, n = 299, s = \frac{10}{\sqrt{299}} = 0.5783$

If 299 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 0.7 millimeters

This is the pvalue of Z when X = 170 + 0.7 = 170.7 subtracted by the pvalue of Z when X = 170 - 0.7 = 169.3. So

X = 170.7

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{170.7 - 170}{0.5783}$

Z = 1.21

Z = 1.21 has a pvalue of 0.8869

X = 169.3

$Z = \frac{X - \mu}{s}$

$Z = \frac{169.3 - 170}{0.5783}$

Z = -1.21

Z = -1.21 has a pvalue of 0.1131

0.8869 - 0.1131 = 0.7738

77.38% probability that the mean length of the sample rods would differ from the population mean by less than 0.7 millimeters

Answer from: Quest

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn

Answer from: Quest

it is b)

hope this : )

step-by-step explanation:

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