Question

(6 points) Wasserman (1989) studied a process for the manufacturing of steel bolts. Historically, these bolts have a mean thickness of 10.0 mm and a standard deviation of 1.6 mm. In a quality check the engineer has a sample of 5 randomly selected and measured. (a) (2 points) Assuming a near normal distribution what are the mean and standard error (standard deviation of the sample mean) of these quality checks? (b) (2 points) A recent sample of five wafers yielded a sample mean of 10.4 mm. Find the probability of observing such a mean of something larger based on the historic mean and standard deviation. (c) (2 points) 90% of the means taken from samples of 5 should be smaller than what value?

Answer 1

Answer:

a) Mean = 10, standard error = 0.7155

b) 28.77% probability of observing such a mean of something larger based on the historic mean and standard deviation.

c) 12.048 mm

Step-by-step explanation:

To solve this question, we have 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 10, \sigma = 1.6, n = 5, s = \frac{1.6}{\sqrt{5}} = 0.7155$

(a) (2 points) Assuming a near normal distribution what are the mean and standard error (standard deviation of the sample mean) of these quality checks?

Mean = 10, standard error = 0.7155

(b) (2 points) A recent sample of five wafers yielded a sample mean of 10.4 mm. Find the probability of observing such a mean of something larger based on the historic mean and standard deviation.

This is 1 subtracted by the pvalue of Z when Z = 10.4. So

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

By the Central Limit Theorem

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

$Z = \frac{10.4 - 10}{0.7155}$

$Z = 0.56$

$Z = 0.56$ has a pvalue of 0.7123.

1 - 0.7123 = 0.2877

28.77% probability of observing such a mean of something larger based on the historic mean and standard deviation.

(c) (2 points) 90% of the means taken from samples of 5 should be smaller than what value?

This is the value of X when Z has a pvalue of 0.90. So it is X when Z = 1.28.

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

$1.28 = \frac{X - 10}{1.6}$

$X - 10 = 1.6*1.28$

$X = 12.048$