SOLUTION: Very difficult problem! 1. For a particular sample of 50 scores on a psychology exam, the following results were obtained. First quartile = 52 Third quartile = 84 Standard devia

Question 227164: Very difficult problem!
1. For a particular sample of 50 scores on a psychology exam, the following results were obtained.
First quartile = 52 Third quartile = 84 Standard deviation = 10 Range = 66
Mean = 79 Median = 78 Mode = 81 Midrange = 65
Answer each of the following; show all work.
- What score was earned by more students than any other score? Why?
- What was the highest score earned on the exam?
- What was the lowest score earned on the exam?
- According to Chebyshev's Theorem, how many students scored between 59 and 99?
- Assume that the distribution is normal. Based on the Empirical Rule, how many students scored between 69 and 89? (Points: 10)

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1. For a particular sample of 50 scores on a psychology exam, the following results were obtained.
First quartile = 52 Third quartile = 84 Standard deviation = 10 Range = 66
Mean = 79 Median = 78 Mode = 81 Midrange = 65
Answer each of the following; show all work.
- What score was earned by more students than any other score? Why?
Ans: mode = 81
-----------------------------
- What was the highest score earned on the exam?
Ans: median + (1/2)range = 78 + 33 = 111
-----------------------------
- What was the lowest score earned on the exam?
Ans: median - (1/2)range = 78-33 = 45
----------------------------------------
- According to Chebyshev's Theorem, how many students scored between 59 and 99?
Ans: That is the number within 2 standard deviations which is 1-(1/2)2 = 3/4
or 75% of the students
--
(3/4)*50 = 37.5 or, rounding up, 38 students
--------------------------------------------------
- Assume that the distribution is normal. Based on the Empirical Rule, how many students scored between 69 and 89? (Points: 10)
Ans: 0.9545*50 = 47.72
Rounding up you get 48 students
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Cheers,
Stan H.

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