Question

ribution by taking a transformation of the data. For the data shown below, take the natural logarithm of the data and answer the following questions 79725, 12862, 18022, 76712, 256440, 14013, 46083, 6808, 85781, 1251 6081, 50397, 11020, 13633, 1064, 496433, 25308, 6616, 11210, 13900 (a) Determine the mean, median, and standard deviation. median (b) Find the 25th, 50th, and 75th percentiles. 25th (c) Plot the data in a boxplot.

Answer 1

79725 , 12862 , 18022 , 76712 , 256440 , 14013 , 46083 , 6808 , 85781 , 1251 , 6081 , 50397 , 11020 , 13633 , 1064 , 496433 , 25308 , 6616 , 11210 , 13900

.

(a)

Mean is calculated by adding all the 20 values and dividing the sum by 20.

To find out the median, the 20 values are arranged in increasing order at first and the average of 10th and 11th value is the median.

Variance is calculated by adding the squares of all the values an dividing the sum by 20, and then subtracting the square of the mean from it. Standard deviation is the square root of variance.

mean = 61667.95

median = 13956.5

standard deviation = 114563.1299

.

(b)

To find out the 25th percentile or the lower quartile, the 20 values are arranged in increasing order at first and the average of 5th and 6th value is the lower quartie.

The 50th percentile is the median.

To find out the 75th percentile or the upper quartile, the 20 values are arranged in increasing order at first and the average of 15th and 16th value is the upper quartile.

25th percentile = 8914

50th percentile = 13956.5

75th percentile = 63554.5

.

(c)

.

(d)

Logarithmic transformation :-

11.286338 , 9.462033 , 9.799349 , 11.247813 , 12.454650 , 9.547741 , 10.738199 , 8.825854 , 11.359553 , 7.131699 , 8.712924 , 10.827687 , 9.307467 , 9.520249 , 6.969791 , 13.115204 , 10.138876 , 8.797246 , 9.324562 , 9.539644

No

Logarithmic transformation does not result in somewhat symmetric distribution.

From the box plot it can be seen that lower quartile is much nearer to the median than the upper quartile. Also from the histogram it is evident that the data is positively skewed.