The SAT placement test for admission to American universities had an average of μ = 1026 and a standard deviation σ = 209. The distribution of marks is approximately normal.

a) Ramon had a grade of 1100. If the notes have a normal distribution, to what percentile does Ramon's note correspond?

b) Now consider the average X¯ of the scores of 70 randomly selected students. If X¯ = 1100, at what percentile of the sampling distribution of X¯ does this result correspond?

c) Which of your two calculations, a) or b), is less precise since SAT notes do not follow exactly a normal law?

Here is my approach:

a) I calculated z-score = 0.354 then I found that the proportion correspond is 0.6368. So what is the percentile correspond? 63.68 or 64?

b) I obtained z-score =2.96 so the proportion is 0.9985. Then what is the percentile?

  • $\begingroup$ Hi @Simon Simmons - I just answered your question. Let me know if you need additional clarification. $\endgroup$ – 324 Dec 11 '19 at 15:58

(a) Computation using R statistical software.

pnorm(1100, 1026, 209)
[1] 0.6383557

In this testing context, percentile is defined as 'percent below', so I'd say 63.83%, often rounded to 64% (except that rounding to integers is unusual for results above 90%).

(b) Computation in R: I'd say 99.85%.

pnorm(1100, 1026, 209/sqrt(70))
[1] 0.9984734

Note: If you are getting probabilities from printed tables of the normal CDF, then you may need to round in order to use the tables. Thus it is possible for you to get slightly different results than the exact ones I got from R.

(c) I suppose you are expected to say that (b) is more accurate because the Central Limit Theorem will tend to make the average of 70 scores more nearly normal than an individual score. (A quibble in this practical application is that fitting a normal model is more difficult in the tails than near the center of the distribution.)


If you are less familiar with R, you can also use Excel of a TI-84 calculator.

(a) Use the normalCDF function which takes the parameters normalCDF(lower bound, upper bound, mean, standard deviation)

In your case, your lower bound is technically $- \infty$, but entering a large, arbitrary negative number will do. Your upper bound is the grade of 1100. This would look like the following:

normalCDF(-1000000, 1100, 1026, 209) which yields 0.6383555986 (so approximately the 64th percentile... you could keep it as the 63.68th percentile if you really wanted to be exact.)

(b) Perform the same calculation as in (a), but change your standard deviation to $\frac{209}{\sqrt{70}}$.

(c) For Part C, how many students are there in part (a). Assuming Normality, possibly part (b) would be better since you know you are getting the average of 70 scores. If part (a) is bringing in a larger sample size $n$, then it would be more normal.

Typically, the larger the $n$, the more Normal.


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