Mean & Standard Deviation of a Random Variable

[asanz89]

I'm working on my homework and am having some trouble with the following question.. I was hoping someone could help me out!!

On a test that is known to produce scores which fit the Standard Normal Distribution,

a. what is the percentile of a person with a score of 79 on a test which supposedly has a mean of 87 and a standard deviation of 6?
b. what scores lie at the 25th and 75th percentile?
c. what proportion of test-takes would be expected to get a score of 97 or higher?

So for a,
y=79
mean=87
standard deviation=6

z=y-mean/s.d.
z=79-87/6
z=-1.33

Then I used the normal distribution table to find 1.33=0.4082.. 0.5-0.4082=0.0918x100 = 9th percentile

I'm completely lost for b and c, if anyone could please help me understand it'd be greatly appreciated!!

 

[galactus]

a. what is the percentile of a person with a score of 79 on a test which supposedly has a mean of 87 and a standard deviation of 6?

b. what scores lie at the 25th and 75th percentile?

Look up .25 in the body of the table and the corresponding z score is -.675

Look up .75 in the body of the table and the corresponding z score is .675

\(\displaystyle -.675=\frac{x-87}{6}\)

\(\displaystyle .675=\frac{x-87}{6}\)

Solve for x to find the two scores.

c. what proportion of test-takes would be expected to get a score of 97 or higher?

\(\displaystyle z=\frac{97-87}{6}=1.67\)

Look up the proportion in the table and subtract from 1.

 

[asanz89]

Thanks for your help!

Just a quick clarification -

for b,

For the 25th percentile - I did 0.5-0.25 = 0.25, then used the backward normal table (critical values of z). For .25, I got .674

For the 75th percentile - I did 0.5-0.75 = -0.25, so should I use -.674?

.674 = x-87/6
.674(6) = x-87
.674(6)+87 = x
91.04 = x

-.674 = x-87/6
-.674(6) = x-87
-.674(6)+87 = x
82.96 = x

For c (I'm still having some trouble),

z= 97-87/6
z=1.67

In the table, it gives me .4525, so if I subtract 1-.4525 = .55.. That doesn't seem like an accurate number for the question?

Just want to ensure I'm on the right track! Thanks again!

 

[galactus]

\(\displaystyle -.675=\frac{x-87}{6}\Rightarrow x=83\)

\(\displaystyle .675=\frac{x-87}{6}\Rightarrow x=91\)

part c. You have the table where you have to add .5. Note, if you add .5 to .4525, you get .9525.

The z score for 1.67 is .9525

Since they ask for 97 or higher, subtract from 1

1-.9525=.0475

About 5% will score 97 or higher.

These are the ones in around the 95th percentile. The students who done real well.

 

[asanz89]

Ah, I was confused by the table!

Thanks so much for all of your help!

I'm not even sure if we need to answer the following question, but I'm curious.. The final part just states: if a randomly selected individual gets a score of 97, what would a statistician assume about the population parameters?

I know the population parameters describes a population distribution, through the mean (87) and standard deviation (6). How is that significant for one to assume something about that individual?