In a normal distribution with a mean of 0 and a standard deviation of 1 find the area under the normal curve with lies:

above z= 1.64
below z = -1.96
between z = -1.50 and +1.50

In a normal distribution with a mean of 100 and a standard deviation of 16 find the percentile equivalent of the following scores;
120
95

On a standardized moth aptitude test the norms for males are:
mean = 50 and s = 6
and the norms for females are:
mean = 55 and s = 8

Harry scored at eh 805ile
What was his raw score on the test?
What was his %ile rank on the female norms?
Can someone help simplify this for me?? :confused:

Susan,

What have you tried so far?

The first question just requires you to use the normal distribution tables (aka z-tables) in your textbook to find the area under the normal curve.

The second and third questions are basically the same as the first - you just need to convert the scores into z scores: z = (x - mu)/s

then convert the z scores into percentiles.

Your textbook probably has examples that are basically the same as these questions.

JohnM

I am struggling to understand exactly what is needed. When reading an explanation of standard dev I was able to understand and go step by step. instead of looking at examples. We quit using our text book after the first five chapters on logic of statistics. Now I have a bunch of handouts that I am sorting through.
Thanks

Susan,

For the first example, to find the area under the curve above z=1.64, find z=1.64 in the normal distribution table.

Usually it gives you the area between negative infinity and z. You should get an area of 0.9495. This is the area between negative infinity and 1.64. Since the entire area under the curve is =1, then the area beyond z=1.64 will be 1-.9495 = 0.0505.

In order to convert a z score into a percentile, you just need to know how much of the area lies below it. In this example, 94.95% of the area lies below z=1.64, so it is the 94.95 percentile, or approximately the 95th percentile.

Thanks :)
I had that one correct. I have .3125 for below -1.96

For the next one with the s of 16 and the m of 10 for 120, I have 1.25 89.9%ile..Am I on the right track?

You don't know how much I appreciate you taking your time to help me. Its frustrating to not understand but yet I can't seem to put it down and walk away.

Thanks!!!

I had that one correct. I have .3125 for below -1.96

That one should actually be about 0.025.


For the next one with the s of 16 and the m of 10 for 120, I have 1.25 89.9%ile..Am I on the right track?

Yes, that one is good.

I had that one correct. I have .3125 for below -1.96

That one should actually be about 0.025.


For the next one with the s of 16 and the m of 10 for 120, I have 1.25 89.9%ile..Am I on the right track?

Yes, that one is good.

Could you possibly give me some guidance on the last problem? Are you and instructor?:)

Susan,

No, the site administrator (screen name "quark") and I are just trying to develop a stats help site that will eventually make $$$.:D

Here's some guidance on that problem:

On a standardized moth aptitude test the norms for males are:
mean = 50 and s = 6
and the norms for females are:
mean = 55 and s = 8

Harry scored at the 80%ile
What was his raw score on the test?
find the z score that corresponds to the 80th percentile
use the formula: z = (x-mu)/s --> z = (x - 50)/6
plug in the value you find for z, then solve for x

What was his %ile rank on the female norms?
take his raw score (x) from the previous question, plug into:
z = (x - 55)/8
solve for z
determine the area below z

Looking on the chart looks like 84 is the raw score? You have really helped me. I will forward this on to classmates.

For the 80th percentile, z = 0.842

so then

.842 = (x - 50)/6

and solve for x to get the raw score

Thanks for passing this on to other people.

John

Looking on the chart looks like 84 is the raw score? You have really helped me. I will forward this on to classmates.

So John, is that correct???????

Susan,

No - solving for x here:

.842 = (x - 50)/6

(.842*6) = x - 50

(.842*6) + 50 = x

x = 55.05

Susan,

No - solving for x here:

.842 = (x - 50)/6

(.842*6) = x - 50

(.842*6) + 50 = x

x = 55.05

Thanks John

For the 80th percentile, z = 0.842

so then

.842 = (x - 50)/6

and solve for x to get the raw score

Thanks for passing this on to other people.

John

I have these exact same questions due today at 5:30 and I need help setting them up I am lost as to what goes where

christina... dont fret... this is basically the z formula backwards to find the percentile or rather (x)

so here is your problem:

z=.842
mean=50
sd (standard deviation) is 6

which is basically z=x-mean/sd
so since you dont have x it would be

.842=x-50/6

1.divide 6 into the bottom and top so it cancels out and multiply it to the other side

50/6
__ and 6*.842=x-50
6 /6

the sixs cancel out

2. you should not have
6 *.842=x-50
3. now multiply .842* 6 and you hould get 5.052
4. add 50 to the right (so it cancels out
and add 50 to the left so it is 50+5.052
5. your x = 55.052 or 55 (50+50.052)

that should be right