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I have this question which i am so confused how to solve it

this is on normal curve. The following table is used to determine the student’s grades:

Grade       Z-score 
A above     1.5 
B           0.5-1.5 
C           -0.5 to 0.5 
D           -1.5 to -0.5 
F           Below-1.5 

Using this system, what percent of As Bs, Cs Ds and Fs are expected with a normal distribution of scores?

Now, I know the distribution is distributed in -2.1% - 13.4% - 34.1% (on both sides). It would have been easier if i could imagine that mean is 0 and sd is 1. but the above distribution doesn't match if i assume SD is 1 since the gap between 0.5 is 1 and then its 1 and so on (not sure if i am making any sense, I am just trying to think).

I am unable to put a finger how to go about this problem.

Thanks

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  • $\begingroup$ Z scores do have mean 0 and sd 1. Are you using tables? $\endgroup$ – Glen_b Sep 20 '16 at 4:20
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Use this PDF. It shows area under the gaussian (or population percentage for a normal distribution) for different z-scores.

You have a confidence interval of 0, so use the first column only!

To find the number of students who will get an F, you need to find the area under the gaussian between the left-most point of the distribution and -1.5 z-score. So, in the 'z' column, go down to the value z=-1.5. The value is .0668, so 6.68% of people will receive an F.

To find the number of students who will get a D, you need to find the area under the gaussian between the -1.5 z-score and the -.5 z-score. So, in the 'z' column, go down to the value z=-1.5. The value is .0668. Then, go down the value z=-.5. The value is .0668. Both these areas are from their respective z-scores to the leftmost point on the distribution. So, we subtract the z=-1.5 value from the z--.5 value to get the area between the two. .3085-.0668 = .2417, so 24.17% will get a D.

Rinse, lather repeat. (Hint: The distribution is symmetrical, so you shouldn't need to explicitly find the areas for 'A' and 'B')

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