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April 21, 2014

April 21, 2014

Posted by **Dawn** on Saturday, September 29, 2012 at 8:35am.

First quartile = 57 Third quartile = 72 Standard deviation = 9 Range = 52

Mean = 72 Median = 68 Mode = 70 Midrange = 57

Answer each of the following:

I. What score was earned by more students than any other score? Why?

II. What was the highest score earned on the exam?

III. What was the lowest score earned on the exam?

IV. According to Chebyshev's Theorem, how many students scored between 54 and 90?

V. Assume that the distribution is normal. Based on the Empirical Rule, how many students scored between 54 and 90?

Please show all work!

- statistics -
**MathGuru**, Saturday, September 29, 2012 at 9:37amI'll give you a few hints.

For the first three questions, you should be able to look at the data given and determine the values. (For example, remember that the mode is the most frequently occurring score in a distribution.)

For the last two questions, see below.

Chebyshev's Theorem says:

1. Within two standard deviations of the mean, you will find at least 75% of the data.

2. Within three standard deviations of the mean, you will find at least 89% of the data.

Empirical Rule says:

1. Within two standard deviations of the mean, you will find about 95% of the data.

2. Within three standard deviations of the mean, you will find about 99.7% of the data.

From the data given:

Mean = 72, Standard deviation = 9

Therefore, between 54 and 90 fall within 2 standard deviations about the mean. (72 - 18 = 54; 72 + 18 = 90)

To answer your last two questions, take the appropriate percentages and multiply by the sample size of 63 to get the number of students between 54 and 90.

I hope this will help.

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