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 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?
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.