Question

A quality control manager at an auto plant measures the paint thickness on newly painted cars. A certain part that they paint has a mean thickness of 2 mm with a standard deviation of 0.8 mm. The manager takes a daily random sample of thickness measurements from 100 points of the samples. What is the probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value?

Answer 1

Answer:

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

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 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 $s = \frac{\sigma}{\sqrt{n}}$.

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

In this problem, we have that:

$\mu = 2, \sigma = 0.8, n = 100, s = \frac{0.8}{\sqrt{100}} = 0.08$

What is the probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value?

This is the pvalue of Z when X = 2 + 0.1 = 2.1 subtracted by the pvalue of Z when X = 2 - 0.1 = 1.9. So

X = 2.1

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

By the Central Limit Theorem

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

$Z = \frac{2.1 - 2}{0.08}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

X = 1.9

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

$Z = \frac{1.9 - 2}{0.08}$

$Z = -1.25$

$Z = -1.25$ has a pvalue of 0.1056

0.8944 - 0.1056 = 0.7888

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

Answer 2

There is 78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value.

For normally distributed random variable X,

Given that,  mean $\mu=2$ , standard deviation $\sigma=0.8$,number of sample,$n=100$

 $s=\frac{0.8}{\sqrt{100} }=0.08$

For random variable, $X=2+0.1=2.1$

By central limit theorem, $Z=\frac{X-\mu}{s}=\frac{2.1-2}{0.08} =1.25$

From Z- value table,

$Z=1.25$ has a p-value of 0.8944.

For random variable, $X=2-0.1=1.9$

By central limit theorem, $Z=\frac{X-\mu}{s}=\frac{1.9-2}{0.08} =-1.25$

From Z- value table,

$Z=-1.25$ has a p-value of 0.1056

So that,  $p=0.8944-0.1056=0.7888$

Hence, There is 78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value.

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