Question

The article "Monte Carlo Simulation—Tool for Better Understanding of LRFD" (J. Structural Engr., 1993: 1586–1599) suggests that yield strength (ksi) for A36 grade steel is normally distributed with 43 and 4.5. a. What is the probability that yield strength is at most 40? Greater than 60? b. What yield strength value separates the strongest 75% from the others?

Answer 1

Answer:

a) $P(X<40)=P(\frac{X-\mu}{\sigma}<\frac{40-\mu}{\sigma})=P(Z<\frac{40-43}{4.5})=P(z<-0.667)$

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

$P(z<-0.667)=0.252$

b) $P(X>40)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-43}{4.5})=P(z>3.78)$

And we can find this probability using the complement rule and normal standard distribution or excel and we got:

$P(z>3.78)=1-P(Z<3.78)=1-0.999922=0.0000784$

c) $z=0.674<\frac{a-43}{4.5}$

And if we solve for a we got

$a=43 +0.674*4.5=46.033$

So the value of height that separates the bottom 75% of data from the top 25% is 46.033.  

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".  

Solution to the problem

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:

$X \sim N(43,4.5)$  

Where $\mu=43$ and $\sigma=4.5$

Part a

We are interested on this probability

$P(X<40)$

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:

$P(X<40)=P(\frac{X-\mu}{\sigma}<\frac{40-\mu}{\sigma})=P(Z<\frac{40-43}{4.5})=P(z<-0.667)$

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

$P(z<-0.667)=0.252$

Part b

$P(X>40)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-43}{4.5})=P(z>3.78)$

And we can find this probability using the complement rule and normal standard distribution or excel and we got:

$P(z>3.78)=1-P(Z<3.78)=1-0.999922=0.0000784$

Part c

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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.75$  

$P(z<\frac{a-\mu}{\sigma})=0.75$

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

$z=0.674<\frac{a-43}{4.5}$

And if we solve for a we got

$a=43 +0.674*4.5=46.033$

So the value of height that separates the bottom 75% of data from the top 25% is 46.033.