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Unformatted text preview: deviations is unchanged. See p. 16, bottom, of the text.
5. The sum of the xi is
100
i=1 xi = 100
i=5 xi fi = (0 · 6) + (1 · 7) + (2 · 19) + (3 · 14) + (4 · 4) = 103. This is because, if we assume that the xi are listed so that all of the 0’s come ﬁrst, followed by all
the 1’s, followed by all the 2’s, etc, then (0 · 6) gives the sum of the ﬁrst 6 Xi , (1 · 7) gives the sum
1
¯
of the next 7 Xi , etc. Thus the sample mean is x = 50 · 103 = 2.06. For the sample median, since n = 50 is even, the sample median will be the average of the 25th and
26th values. But from the table we see that the 14th through 32nd values are all 2, so, in particular,
the 25th and 26th values are both 2, and so the sample median is 2.
50
1
(xi )2 − 50 · x2 .
¯
To calculate the sample variance, we use the alternate formula s2 = 49
i=1
50
2 = 50 · 103 · 103 = 103 · 2.06 = 212.18. For
2 , by the same reasoning used
Note that 50 · x
¯
i=1 (xi )
50
50
in our calculation of x, this can be calculated as (0 · 6) + (1 · 7) + (4 · 19) + (9 · 14) + (16 · 4) = 273,
¯
√
1
so ﬁnally, s2 = 49 (273 − 212.18) = 1.241224, and...
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This document was uploaded on 02/15/2014 for the course MATH 231 at Lehigh University .
 Fall '11
 DANIELCONUS
 Statistics

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