Question

A factory produces plate glass with a mean thickness of 4 mm and a standard deviation of 1.1 mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 3.83 mm? . Round your answers to 5 decimal places.

Answer 1

Answer:

$P(\bar X<3.83)=0.06117$

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:

$X \sim N(4,1.1)$  

Where $\mu=4$ and $\sigma=1.1$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case  $\bar X \sim N(4,\frac{1.1}{\sqrt{100}})$

2) Solution to the problem

We are interested on this probability

$P(\bar X<3.83)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X<3.83)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{3.83-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(Z<\frac{3.83-4}{\frac{1.1}{\sqrt{100}}})=P(Z<-1.545)$

And in order to find this probability we can find tables for the normal standard distribution, excel or a calculator.  

$P(Z<-1.545)=0.06117$

And the excel formula to calculate it would be:

"=NORM.DIST(-1.545,0,1,TRUE)"

Answer 2

The probability is 49.20%

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation.\\\\For\ a\ sample\ size\ n:\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }$

Given that n = 100, μ = 4 mm, σ = 1.1 mm

For x < 3.83 mm:

$z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\z=\frac{3.83-4}{1.1/\sqrt{100} } =-0.0187$

P(x < 3.83) = P(z < -0.0187) = 0.4920 = 49.20%

From the normal distribution table, the probability that the average thickness of the 100 sheets is less than 3.83 mm is 49.20%

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