Thread: Help - questions re: Empirical Rule/Chebyshev

1. Help - questions re: Empirical Rule/Chebyshev

Hello:

Does anyone know how to approach these problems? Any help would be greatly appreciated.

The mean IQ of 200 patients in a psychiatric hospital is 91, with a variance of 16, and the distribution is highly negatively skewed.

a) Between what two IQ scores would we expect to find at least 160 of the patients falling? ANS: 82 to 100

b) Suppose the distribution were mound-shaped and symmetrical (with the same mean and standard deviation). Approximately, how many patients would have IQ scores at or below the 5th percentile or above a score of 97? ANS: 28.5 or 29

2. question (a)
You need to take Chebyshev's Theorem, and figure out between what 2 scores you would find 160/200 (80%) of the patients

i.e., 1 - (1/k^2) = 80% --> solve for k, and this will be the number of standard deviations

question (b)
Use the Empirical Rule (i.e., 68% within +/- 1 std dev, 95% within +/-2, etc.) to determine the % of patients' scores between the given boundaries
i.e., how many standard deviations from the mean is the 5th percentile, how many st dev from the mean is a score of 97?

use the formula z = (x - mu)/s

3. Hi John,
I was able to figure out part A for this question
but for Part B i am having some difficulty still

to find the number of patients who have IQ scores above a score of 97 this is what I did:

Z = 97-91/4 = 1.5 std. dev. from the mean
1.5 std. dev. also means 81.5 percentile
so p = 1 - 0.815 = 0.185
so 0.185 x 91 (mean) = 16.835 patients have IQ scores above a score of 97

but I am having trouble trying to figure out how many patients have IQ scores at or below the 5th percentile..how do I find out what score corresponds to the 5th percentile?

4. Look up the z score (in the normal distribution table) that corresponds to the 5th percentile, then plug it into:

z = (x - 91)/4

and solve for x, which will be the score you're looking for.

5. hmm im not quite sure what you mean by look up the z-score
would i be correct in saying that a score of 85.705 corresponds to the 5th percentile?

if i already have the percentile why cant i just multiply that by the mean and get my answer? (kind of like what i did for people who score over 97)

6. Actually, you would expect 5% of the scores to be below the 5th percentile.

Then you find the % of scores above 97:

z = (97-91)/4 = 1.5

P(Z > 1.5) = .067

So, add .067 + .05 = .117

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