Question

In: Statistics and Probability

Annual sales, in millions of dollars, for 21 pharmaceutical companies follow. 8,072 1,319 1,798 8,524 2,361...

Annual sales, in millions of dollars, for 21 pharmaceutical companies follow. 8,072 1,319 1,798 8,524 2,361 10,956 585 14,138 6,194 1,776 2,705 1,302 10,078 7,179 3,859 4,167 709 2,042 3,507 5,562 7,974 Provide a five-number summary. Smallest value First quartile Median Third quartile Largest value Compute the lower and upper limits. Lower limit Upper limit Do the data contain any outliers? Johnson & Johnson's sales are the largest on the list at $14,138 million. Suppose a data entry error (a transposition) had been made and the sales had been entered as $41,138 million. Would the method of detecting outliers in part (c) identify this problem and allow for correction of the data entry error? Which of the following box plots accurately displays the data set?

Solutions

Expert Solution

five-number summary.

Smallest value = 585

First quartile :

The first quartile (or lower quartile or 25th percentile) is the median of the bottom half of the numbers. So, to find the first quartile, we need to place the numbers in value order and find the bottom half.

585   709   1302   1319   1776   1798   2042   2361   2705   3507   3859   4167   5562   6194   7179   7974   8072   8524   10078   10956   14138   

So, the bottom half is

585   709   1302   1319   1776   1798   2042   2361   2705   3507   

The median of these numbers is 1787.

First quartile = 1787

Median :

The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

585   709   1302   1319   1776   1798   2042   2361   2705   3507   3859   4167   5562   6194   7179   7974   8072   8524   10078   10956   14138   

So, the median is 3859 .

Third quartile :

The third quartile (or upper quartile or 75th percentile) is the median of the upper half of the numbers. So, to find the third quartile, we need to place the numbers in value order and find the upper half.

585   709   1302   1319   1776   1798   2042   2361   2705   3507   3859   4167   5562   6194   7179   7974   8072   8524   10078   10956   14138   

So, the upper half is

4167   5562   6194   7179   7974   8072   8524   10078   10956   14138   

The median of these numbers is 8023.

Third quartile = 8023

maximum value = 14138

IQR =  8023 - 1787 = 6236.

Lower limit = Q1 - 1.5 ( IQR) = 1787 - 1.5 ( 6236 ) = - 7567

Upper limit = Q3 + 1.5 ( IQR ) = 8023 + 1.5 ( 6236 ) = 17377

There are no outliers.

largest value on the data is 14138 and if due to data entry it is entered as 41138, then the values of quartiles will not be affected. Because there is no efffect of extreme values on the quartiles. And hence the method of detecting outlier will work here. IT will find the outlier and correction will be made.


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