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Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.5. (Round your answers to four decimal places.)(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 10 pins is at least 51

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Answer:

0.0174 = 1.74% probability that the sample mean hardness for a random sample of 10 pins is at least 51

Explanation:

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

Normal probability distribution

When the distribution is normal, we use the z-score formula.

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


Z = (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 normally distributed random variable X, with mean
\mu and standard deviation
\sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
\mu and standard deviation
s = (\sigma)/(√(n)).

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:


\mu = 50, \sigma = 1.5, n = 10, s = (1.5)/(√(10)) = 0.4743

What is the probability that the sample mean hardness for a random sample of 10 pins is at least 51

This is 1 subtracted by the pvalue of Z when X = 51. So


Z = (X - \mu)/(\sigma)

By the Central Limit Theorem


Z = (X - \mu)/(s)


Z = (51 - 50)/(0.4743)


Z = 2.11


Z = 2.11 has a pvalue of 0.9826

1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean hardness for a random sample of 10 pins is at least 51

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