Question

From 6.1-6. We have the following 20 data points: 0.076357441, 0.165254709, 0.054148344, 0.015440066, 0.070112429. О.063968165, 0.013389992, 0.033028765, 0.040745564, 0.196315217, 0.021880427, 0.030107953, 0.049859409, 0.017333472, 0.053670154, 0.048490605, 0.010339937, 0.096573528, 0.006190141, 0.041041448

Answer 1

(1) The five-number summary can be computed as follows:

Sorting the data,

0.006190141, 0.0103399347, 0.013389992, 0.015440066, 0.017333472,0.021880427, 0.030107953, 0.033028765, 0.040745564, 0.041041448, 0.048490605, 0.049859409, 0.053670154, 0.054148344, 0.063968165,0.070112429,0.076357441, 0.096573528, 0.165254709, 0.196315217

1.We first compute the median, which is the average of the 2 middle no.s (since the series is even with 20 observations, we choose the 10th and 11th observation,)

Median = 0.0447660265

2.First Quartile = Middle no. of the smallest number and median of the dataset = Q₁

Now Q₁ = Middle no. of 0.006190141 ( Smallest), 0.0103399347, 0.013389992, 0.015440066,

0.017333472, 0.021880427,0.030107953, 0.033028765, 0.040745564, 0.0447660265 (Median)

= =  0.0196069495

3. Similarly, Third Quartile = Middle no. of the median and largest number of the dataset = Q₁

Now Q₃ = 0.067040297

4. Minimum = Minimum of the 20 data points = 0.006190141

5. Maximum = Maximum of the 20 data points = 0.196315217

(2) The interquartile range (IQR) is computed using the formula:

= 0.067040297 - 0.0196069495

IQR = 0.0474333475

The inner and outer fences gives the extreme values in the dataset. They are computed as follows:

Lower fence = Q₁ – 1.5 ( IQR )

= 0.0196069495 - 1.5 (0.0474333475)

Lower fence = -0.05154307175

Upper fence = Q₃ + 1.5 ( IQR )

= 0.067040297 + 1.5 (0.0474333475)

Upper fence = 0.13819031825