Question

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 42 and σ = 5.5.

(a) What is the probability that yield strength is at most 40? Greater than 64? (Round your answers to four decimal places.)
at most 40
greater than 64
(b) What yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.)
ksi

Answer 1

a)
`P( X <40) =P(Z< (40-42)/(5.5)) =P(Z<-0.363)=0.3583`

We want this probability:

`P( X >64)`

And using the z score formula given by:

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

We got:

`P( X >64) =P(Z> (64-42)/(5.5)) =P(Z>4)=0.0000316`

b) For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.25` (a)

`P(X<a)=0.75` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.75`

`P(z<(a-\mu)/(\sigma))=0.75`

But we know which value of z satisfy the previous equation so then we can do this:

`z=0.674<(a-42)/(5.5)`

And if we solve for a we got

`a=42 +0.674*5.5=45.707`

So the value of height that separates the bottom 75% of data from the top 25% is 45.707.

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

`X \sim N(42,25.5)`

Where
`\mu=42` and
`\sigma=5.5`

And we want this probability:

`P( X <40)`

And using the z score formula given by:

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

We got:

`P( X <40) =P(Z< (40-42)/(5.5)) =P(Z<-0.363)=0.3583`

We want this probability:

`P( X >64)`

And using the z score formula given by:

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

We got:

`P( X >64) =P(Z> (64-42)/(5.5)) =P(Z>4)=0.0000316`

Part b

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.25` (a)

`P(X<a)=0.75` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.75`

`P(z<(a-\mu)/(\sigma))=0.75`

But we know which value of z satisfy the previous equation so then we can do this:

`z=0.674<(a-42)/(5.5)`

And if we solve for a we got

`a=42 +0.674*5.5=45.707`

So the value of height that separates the bottom 75% of data from the top 25% is 45.707.