Question

1 The following data represents the number of innings pitched by the ERA leaders for the past few years 23, 30, 20, 27, 44, 26, 35, 20, 29. 29, 25, 15, 18, 27, 19, 22, 12, 26. 34. 15. 27, 35, 26, ald43 35, 14, 24, 12, 23. 31, 40. 35, 38, 57, 22. 42, 24, 21, 27, 33 1. Find the value that corresponds to the 70th percentile. 2. Identify all the outiiers 3. Construct a boxplot for the data and comment on the shape of the distribution.

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Answer #1

(1) I have solved this question using M.S.Excel -2016.

Firstly put all the data in excel.To get the 70th percentile value use the "=PERCENTILE.EXC( Array, k)" function in excel. In array give the location of your data and in the place of k put 0.7.

B41 MT v/ ? -PERCENTILE.EXC(B1:840, 0.7) 40 41 70th percentile 42 57 32.4

(2) To identify the outliers we will firstly find the Minimum value, 1st quartile value, Median , 3rd quartile value and maximum value.

Here I have used ="QUARTILE(array, k) function. In the place of array give the range of the data and for in the place of k, for minimum value give 0, for 1st quartile give 1, for Median give 2, for 3rd quartile give 3 and for maximum value give 4.

Now we will find Inter quartile range(IQR) = Q3 - Q1

IQR = 33.75 - 22 = 11.75

Lower Fence = Q1 - 1.5(IQR)

= 22 - (1.5\times11.75) = 4.375

Upper Fence = Q3 + 1.5(IQR)

= 33.75 + (1.5\times 11.75) = 51.37

Hence, any value within the data set lie outside the limit ( 4.375 , 51.37) will be consider as an Outlier.

In our data set there is only one data point, 57 that is lying outside the upper fence limit.

So, 57 is an outlier.

(3)

The median of the distribution = 26.5

Mean = Average of the data set = 27.625

Mode = 27

Here we can see that mean, median and mode are approximately equal to each other. Hence we can say that the shape of the distribution is approximately Normal.

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