Z= X - mean / SD

Z = X - 74 / 7

I don't know where to put 12%.....

is it Z = 0.12 - 74 / 7 ?

Z = -10.554 and I can't find it on the z table.... please help... thanks...

- Thread starter trialqw
- Start date

Z= X - mean / SD

Z = X - 74 / 7

I don't know where to put 12%.....

is it Z = 0.12 - 74 / 7 ?

Z = -10.554 and I can't find it on the z table.... please help... thanks...

Think of a bell curve (the normal distribution). The area that is under the curve will add up to 1.

The middle of that bell is the mean (i.e. 74). This means that the left half covers 50% of the students, and the right half covers the 50% scoring better than the average. But you don't want this. You want to find the actual point such that to the right of the point, you'll have 12% of the students; and to the left of the point, you'll have the other 88%.

Are you with me so far? If so, then what you want to do to find this point is use the cdf...that is, you want to use the integral of the pdf for the normal distribution. The cdf will tell you the point at which you have 88% of the students on the left side. Just solve for x at 0.88.

Hope this helps.

___(88% of the class that did not get an "A")____()_______(12% of the class that got an A)_____

So X is the () ?

Our prof didn't teach us the cdf yet... and I dont even know what cdf stands for.... I googled it and it popped out Cumulative distribution function and I am 100% positive that our prof did not teach that yet...

So I guess I am screwed?

..............................|

..............................|

50% less than ave ....|.........50% that have higher than average

..............................|

..............................|

___________________|________________(? point what i am trying to find)__

..............................74 = mean

I am so lost at trying to find (that point)...............

After researching I reached the point that I should use invNorm in the calculator...

since 88% is to the left of the X, I would then input...

input .88, 74, 7

Can someone help me because I dont have a calc with invNorm function..... to find X that separates the 88% from 12% ....

NORM.S.INV is used if SD is 1 or 0 or and the mean is 1 or 0....

I think I should have used the NORM.INV instead of NORM.S.INV because me SD is 7 and mean is 74

since 12% is the upper.. ind inv norm is used left to right from the X or point... I would use 88% or .88

NORM INV (0.88 ,74,7)

= 82.22491

still almost the same as the NORM.S.INV. though....

So my answer stands still 82 is the X... Is that correct? plss help I have to study for pathology and Obstetrics.... exam tomorrow...

So my answer stands still 82 is the X... Is that correct? plss help I have to study for pathology and Obstetrics.... exam tomorrow...

Addressing your question though, the answer does appear correct. It looks like you might have gotten the first method wrong though. How do you get the 7/7?

I am from the Philippines.... taking up medicine ~ medical student ~ doctor-to-be... the statistics is part of preventive medicine... oh the 7 / 7 stuff.. It was just part of the algebra part... cancelling etc....

I hope Im correct Im gonna complain to the prof coz he hasn't taught cdf or that INV NORM stufff.... This was just a homework exercise...

okay done Pressed the thanks button Link

oh should this be in the statistics or probability section? lol

I hope Im correct Im gonna complain to the prof coz he hasn't taught cdf or that INV NORM stufff.... This was just a homework exercise...

okay done Pressed the thanks button Link

oh should this be in the statistics or probability section? lol

Last edited by a moderator: