Statistics problem - Continuous random varibles

[anarovira]

Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean value 15.0 kips and standard deviation 1.25 kips.
Compute the following probabilities by standardizing and then using Table A.3.

a) P(X ≤ 15)
b) P(X ≤ 17.5)
c) P(X ≥ 10)
d) P(14 ≤ X ≤ 18)
e) P(|X - 15| ≤ 3)

table: http://www.stat.tamu.edu/~twehrly/651/ztable.pdf

can someone please help? :)

 

[BTP]

Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean value 15.0 kips and standard deviation 1.25 kips.
Compute the following probabilities by standardizing and then using Table A.3.

a) P(X ≤ 15)
b) P(X ≤ 17.5)
c) P(X ≥ 10)
d) P(14 ≤ X ≤ 18)
e) P(|X - 15| ≤ 3)

table: http://www.stat.tamu.edu/~twehrly/651/ztable.pdf

can someone please help? :)

Your z-table gives you the probability to the left of z-value of the distribution Z~N(0,1). In other words let Z~N(0,1), μ=mean=0 and σ=std=1. Now for a given z value the table returns P(Z<z*).

Now the question is how to work with X~N(μ,σ). To compute the probability P(X<x*)you transform the question into one about Z by Z= (X-μ)/σ. This is called standardizing your normal random variable. Under this transformation we find that: P(X<x*) = P(Z< (x*-μ)/σ) !

Example:

So given your distribution: X~N(15kips, 1.25 kips) compute P(X<15 kips).

Solution:

P(X<x*)= P(Z < (x*-μ)/σ) where in this example x*= 15 kips, μ=15 kips, and σ=1.25 kips

so P(X<15 kips ) = P(Z< (15kips - 15kips)/1.25 kips) = P(Z<0)

Note: (x*-μ)/σ is always dimensionless - this is a way to check you did your calculation correctly

Now we look up Z=0 in the the able and we see the value .5 thus P(Z<0)=.5 so that
P(X<15 kips) = .5

Second Example:

Compute P( 13.75 Kips < X < 15 kips).

It follows that P( 13.75 Kips < X < 15 kips) = P(X <15kips) - P(X <13.75 kips)

You should think about why this is true.

So P( 13.75 Kips < X < 15 kips) = P(X <15kips) - P(X <13.75 kips)= P(Z< (15kips -15kips)/1.25 kips) - P(Z<(13.75 kips -15kips)/1.25 kips) = P(Z<0) - P(Z<-1)

Now P(Z<0)=.5 and P(Z<-1)= .1587 (from the table)

Thus P( 13.75 Kips < X < 15 kips) = .5 -.1587 =.343------------------------------------------------------

These two examples should help guide you when solving your problems.

 

[Ray Vickson]

Your z-table gives you the probability to the left of z-value of the distribution Z~N(0,1). In other words let Z~N(0,1), μ=mean=0 and σ=std=1. Now for a given z value the table returns P(Z<z*).

Now the question is how to work with X~N(μ,σ). To compute the probability P(X<x*)you transform the question into one about Z by Z= (X-μ)/σ. This is called standardizing your normal random variable. Under this transformation we find that: P(X<x*) = P(Z< (x*-μ)/σ) !

Example:

So given your distribution: X~N(15kips, 1.25 kips) compute P(X<15 kips).

Solution:

P(X<x*)= P(Z < (x*-μ)/σ) where in this example x*= 15 kips, μ=15 kips, and σ=1.25 kips

so P(X<15 kips ) = P(Z< (15kips - 15kips)/1.25 kips) = P(Z<0)

Note: (x*-μ)/σ is always dimensionless - this is a way to check you did your calculation correctly

Now we look up Z=0 in the the able and we see the value .5 thus P(Z<0)=.5 so that
P(X<15 kips) = .5

Second Example:

Compute P( 13.75 Kips < X < 15 kips).

It follows that P( 13.75 Kips < X < 15 kips) = P(X <15kips) - P(X <13.75 kips)

You should think about why this is true.

So P( 13.75 Kips < X < 15 kips) = P(X <15kips) - P(X <13.75 kips)= P(Z< (15kips -15kips)/1.25 kips) - P(Z<(13.75 kips -15kips)/1.25 kips) = P(Z<0) - P(Z<-1)

Now P(Z<0)=.5 and P(Z<-1)= .1587 (from the table)

Thus P( 13.75 Kips < X < 15 kips) = .5 -.1587 =.343


------------------------------------------------------

These two examples should help guide you when solving your problems.

You did this fellow's homework for him, which is against the Forum rules.

 

[BTP]

Not quite. I explained how the ideas works and did two examples, albeit the first example was problem a), (I did not want to do any complicated examples). My reply is the basic explanation that one would find in any basic statistics text on standardizing a normal random variable. As for the homework problems you are referring to they are about implementing the ideas. There is a difference. If he went for help a teacher would do exactly the same. And which is why I did not do any of the problems listed. In point of fact I left the really crucial point a question for them to figure out - and without which they will not get the rest of the problems - if you are a helper you should know the difference.

 

[anarovira]

Not quite. I explained how the ideas works and did two examples, albeit the first example was problem a), (I did not want to do any complicated examples). My reply is the basic explanation that one would find in any basic statistics text on standardizing a normal random variable. As for the homework problems you are referring to they are about implementing the ideas. There is a difference. If he went for help a teacher would do exactly the same. And which is why I did not do any of the problems listed. In point of fact I left the really crucial point a question for them to figure out - and without which they will not get the rest of the problems - if you are a helper you should know the difference.

Your answer was so helpful, thank you so much! And you indeed didn't do my homework.