Question

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

Answer 1

Answer:

a)$P( X$

We want this probability:

$P( X >64)$

And using the z score formula given by:

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

We got:

$P( X >64) =P(Z> \frac{64-42}{5.5}) =P(Z>4)=0.0000316$

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

$P(X>a)=0.25$   (a)

$P(X$   (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:

$P(X$  

$P(z$

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

$z=0.674$

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.  

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 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:

$P( X$

And using the z score formula given by:

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

We got:

$P( X$

We want this probability:

$P( X >64)$

And using the z score formula given by:

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

We got:

$P( X >64) =P(Z> \frac{64-42}{5.5}) =P(Z>4)=0.0000316$

Part b

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

$P(X>a)=0.25$   (a)

$P(X$   (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:

$P(X$  

$P(z$

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

$z=0.674$

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.