Question

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.1 mm and standard deviation of 0.02 mm.(a) What is the probability that the diameter of the dot exceeds 0.066 mm?(b) What is the probability that the diameter is between 0.07 mm and 0.122 mm?(c) What standard deviation of diameters would be needed so that the probability in part (b) is 0.995?

Answer 1

a)
`P(X<0.066)=P((X-\mu)/(\sigma)<(0.066-\mu)/(\sigma))=P(Z<(0.066-0.1)/(0.02))=P(Z<-1.7)`

And we can find this probability using the normal standard distribution or excel:

`P(Z<-1.7)=0.045`

b)
`P(0.07<X<0.122)=P((0.07-\mu)/(\sigma)<(X-\mu)/(\sigma)<(0.122-\mu)/(\sigma))=P((0.07-0.1)/(0.02)<Z<(0.122-0.1)/(0.02))=P(-1.5<Z<1.1)`

`P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)`

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

`P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)=0.864-0.066=0.798`

c)
`P(0.07<X<0.122)=0.995`

And for this case we can use the z score formula given by:

`z= (x -\mu)/(\sigma)`

We find a value on the normal standard distribution that accumulates 0.0025 of the both tails and the values for this case are
`z= \pm 2.81`

And we can use the z score like this:

`-2.81 = (0.07-0.1)/(\sigma)`

And solving for the deviation we got:

`\sigma= (0.07-0.1)/(-2.81)= 0.0107`

`2.81 = (0.122-0.1)/(\sigma)`

And solving for the deviation we got:

`\sigma= (0.122-0.1)/(2.81)= 0.00783`

So then we can conclude that the standard deviation needs to be between
`0.00783 \leq \sigma \leq 0.0107`

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

`X \sim N(0.1,0.02)`

Where
`\mu=0.1` and
`\sigma=0.02`

We are interested on this probability

`P(X<0.066)`

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

`z=(x-\mu)/(\sigma)`

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

`P(X<0.066)=P((X-\mu)/(\sigma)<(0.066-\mu)/(\sigma))=P(Z<(0.066-0.1)/(0.02))=P(Z<-1.7)`

And we can find this probability using the normal standard distribution or excel:

`P(Z<-1.7)=0.045`

Part b

`P(0.07<X<0.122)=P((0.07-\mu)/(\sigma)<(X-\mu)/(\sigma)<(0.122-\mu)/(\sigma))=P((0.07-0.1)/(0.02)<Z<(0.122-0.1)/(0.02))=P(-1.5<Z<1.1)`

And we can find this probability with this difference:

`P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)`

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

`P(-1.5<Z<1.1)=P(Z<1.1)-P(Z<-1.5)=0.864-0.066=0.798`

Part c

For this case w eneed this condition:

`P(0.07<X<0.122)=0.995`

And for this case we can use the z score formula given by:

`z= (x -\mu)/(\sigma)`

We find a value on the normal standard distribution that accumulates 0.0025 of the both tails and the values for this case are
`z= \pm 2.81`

And we can use the z score like this:

`-2.81 = (0.07-0.1)/(\sigma)`

And solving for the deviation we got:

`\sigma= (0.07-0.1)/(-2.81)= 0.0107`

`2.81 = (0.122-0.1)/(\sigma)`

And solving for the deviation we got:

`\sigma= (0.122-0.1)/(2.81)= 0.00783`

So then we can conclude that the standard deviation needs to be between
`0.00783 \leq \sigma \leq 0.0107`