What Percentile is K in a Normal Distribution?

[FaNgS]

This is a question from the GRE General Test, which I am not able to understand its solution. Maybe someone can help ^^

Question: The variable X is normally distributed. If the values of Xat the 45th,15th and kth percentile are 550, 350 and 450 respectively.

(A) K
(B) 30

Answer Choices:
(A) is greater than (B)
(B) is greater than (A)
(A) and (B) are equal
Can NOT determine the answer from the given infromation.

The answer given is that (B) is greater than (A) (i.e. 30>K)

I don't get it, help please!

 

[Bacle]

I wonder if this is what is going on: Let s:=sigma, Mu:=mean

using the 1-2-3 , aka, 68-95-99.7% rule , 34.3% of data will be within 1 s of the mean

this means that around 15.7% , say 15% will be more than 1s below the mean.

There is another useful rule that 19% of data is within 1/2s of Mu


Use 15% instead of 15.7% . Then approximate:


(350-Mu)/s


To find a general form for s. Then use that to estimate the z-value for

450, and find a bound for it. Then use the fact that the mean is larger

than 550 to bound the z-value of 450, and compare it to 1/2.

Maybe not elegant, but I think it will work.

 

[FaNgS]

thanks for the reply, but is there an easier way to solve it?
you are not allowed to use a calculator in the GRE Test and also the level of mathematics involved is supposedly that of high-school

 

[D H]

You don't need a calculator. All of the points in question are less than the mean. The normal CDF is convex in this interval.

 

[bpet]

Another way to look at it, by drawing a diagram of the normal pdf it's easy to see that the halfway point between the 15th and 45th percentiles divides the corresponding areas into two unequal parts, the greater being nearer the 45th percentile.