Question

A process manufactures ball bearings with diameters that are normally distributed with mean 25.14 mm and standard deviation 0.08 mm. Specifications call for the diameter to be in the interval 2.5 ± 0.01 cm. What proportion of the ball bearings will meet the specification? A. Describe the distribution of ball diameters using proper statistical notation. B. Represent the situation described in the problem graphically. C. Calculate the proportion of ball bearings meeting the specifications

Answer 1

Answer:

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

$X \sim N(2.514 cm,0.008 cm)$  

Where $\mu=2.514$ and $\sigma=0.008$

b) For this case we can see the interval required on the figure attached,

c) $P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)$

And we can find this probability with this difference:

$P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)$

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

$P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307$

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".  

Part a

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

$X \sim N(2.514 cm,0.008 cm)$  

Where $\mu=2.514$ and $\sigma=0.008$

Part b

For this case we can see the interval required on the figure attached,

Part c

We are interested on this probability

$P(2.49<X<2.51)$

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

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

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

$P(2.49<X<2.51)=P(\frac{2.49-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{2.51-\mu}{\sigma})=P(\frac{2.49-2.514}{0.008}<Z<\frac{2.51-2.514}{0.008})=P(-3<z<-0.5)$

And we can find this probability with this difference:

$P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)$

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

$P(-3<z<-0.5)=P(z<-0.5)-P(z<-3)=0.309-0.00135=0.307$