anarovira said:
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.