Question

The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 70 and standard deviation 3. (Rockwell hardness is measured on a continuous scale.)

a. If a specimen is acceptable only if its hardness is between 67 and 75, what is the probability that a randomly chosen specimen has an acceptable hardness?
b. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is independently determined, what is the expected number of acceptable specimens among the ten?
c. What is the probability that at most eight of ten independently selected specimens have a hardness of less than73.84?

Answer 1

a)
`P(67<X<75)=P((67-\mu)/(\sigma)<(X-\mu)/(\sigma)<(75-\mu)/(\sigma))=P((67-70)/(3)<Z<(75-70)/(3))=P(-1<z<1.67)`

And we can find this probability with this difference:

`P(-1<z<1.67)=P(z<1.67-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<1.67)=P(z<1.67)-P(z<-1)=0.953-0.159=0.794`

b) # = 0.794 *10 = 7.94 or approximately 8

c)
`P(X<73.84)=P((X-\mu)/(\sigma)<(73.84-\mu)/(\sigma))=P(Z<(73.84-70)/(3))=P(z<1.28)`

And we can find this probability using excel, calculator or tables:

`P(z<1.28)=0.8997`

And we can approximate the problem with a binomial distribution given by:

`X \sim Bin (n =10, p= 0.8997)`

And we want this probability:

`P(X \leq 8)`

And we can use the complement rule and we got:

`P(X \leq 8)= 1-P(X>8) = 1-[P(X \geq 9)]= 1-[P(X=9)+P(X=10)]`

We find the individual probabilities and we got:

`P(X=9) = (10C9) (0.8997)^9 (1-0.8997)^(10-9)= 0.387`

`P(X=10) = (10C10) (0.8997)^(10) (1-0.8997)^(10-10)= 0.348`

And replacing we got:

`P(X \leq 8)= 1-[0.387+0.348]= 0.265`

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

`X \sim N(70,3)`

Where
`\mu=70` and
`\sigma=3`

We are interested on this probability

`P(67<X<75)`

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(67<X<75)=P((67-\mu)/(\sigma)<(X-\mu)/(\sigma)<(75-\mu)/(\sigma))=P((67-70)/(3)<Z<(75-70)/(3))=P(-1<z<1.67)`

And we can find this probability with this difference:

`P(-1<z<1.67)=P(z<1.67-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<1.67)=P(z<1.67)-P(z<-1)=0.953-0.159=0.794`

Part b

For this case we can multiply the total number by the fraction founded on part a and we got:

# = 0.794 *10 = 7.94 or approximately 8

Part c

For this case we can find the individual probability:

`P(X<73.84)=P((X-\mu)/(\sigma)<(73.84-\mu)/(\sigma))=P(Z<(73.84-70)/(3))=P(z<1.28)`

And we can find this probability using excel, calculator or tables:

`P(z<1.28)=0.8997`

And we can approximate the problem with a binomial distribution given by:

`X \sim Bin (n =10, p= 0.8997)`

And we want this probability:

`P(X \leq 8)`

And we can use the complement rule and we got:

`P(X \leq 8)= 1-P(X>8) = 1-[P(X \geq 9)]= 1-[P(X=9)+P(X=10)]`

We find the individual probabilities and we got:

`P(X=9) = (10C9) (0.8997)^9 (1-0.8997)^(10-9)= 0.387`

`P(X=10) = (10C10) (0.8997)^(10) (1-0.8997)^(10-10)= 0.348`

And replacing we got:

`P(X \leq 8)= 1-[0.387+0.348]= 0.265`