Question

hompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 125 millimeters, and a variance of 49. If a random sample of 41 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by greater than 2.2 millimeters? Round your answer to four decimal places.

Answer 1

Answer:

$P(-2.012<Z<2.012)=P(Z<2.012)-P(Z<-2.012)=0.9779-0.0221=0.9558$

So since we want the probability that "the sample mean would differ from the population mean by greater than 2.2 millimeters". Then we use the complement rule and we got

P=1-0.9558=0.0442

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

$X \sim N(125,49)$  

Where $\mu=125$ and $\sigma=7$

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  $\bar X \sim N(125,\frac{7}{\sqrt{41}}=1.093)$

Solution to the problem

We can begin the problem finding this probability

$P(\mu -2.2<\bar X<\mu+2.2)$

We can use 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:

$P(125-2.2<\bar X<125+2.2)=P(\frac{122.8-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{127.2-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(\frac{122.8-125}{\frac{7}{\sqrt{41}}}<Z<\frac{127.2-125}{\frac{7}{\sqrt{41}}})=P(-2.012<Z<2.012)$

And we can find this probability on this way:

$P(-2.012<Z<2.012)=P(Z<2.012)-P(Z<-2.012)$

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

$P(-2.012<Z<2.012)=P(Z<2.012)-P(Z<-2.012)=0.9779-0.0221=0.9558$

So since we want the probability that "the sample mean would differ from the population mean by greater than 2.2 millimeters". Then we use the complement rule and we got:

P=1-0.9558=0.0442