Question

Shear strength measurements for spot welds have been found to have standard deviation 1 0 pounds per square inch (psi). If 100 test welds are to be measured, what is the approximate probability that the sample mean will be within 1 psi of the true population mean.

Answer 1

Answer:

$P(\mu -1< \bar X <\mu +1)=0.6826$

Step-by-step explanation:

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 Shear strength of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,10)$  

Where $\mu$ and $\sigma=10$

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  $\bar X \sim N(\mu,\frac{10}{\sqrt{100}})$

We are interested on this probability

$P(\mu -1<\bar X<\mu +1)$

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:

$P(\mu -1<\bar X<\mu +1)=P(\frac{\mu- 1-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{\mu +1-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(\frac{\mu -1-\mu}{\frac{10}{\sqrt{100}}}<Z<\frac{\mu +1-\mu}{\frac{10}{\sqrt{100}}})=P(-1<Z<1)$

And we can find this probability on this way:

$P(-1<Z<1)=P(Z<1)-P(Z<-1)$

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

$P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.8413-0.1587=0.6826$