In: Statistics and Probability
Annual sales, in millions of dollars, for 21 pharmaceutical companies follow.
| 8,576 | 1,401 | 1,910 | 9,057 | 2,508 | 11,641 |
| 621 | 14,138 | 6,581 | 1,887 | 2,874 | 1,383 |
| 10,708 | 7,628 | 4,100 | 4,428 | 754 | 2,170 |
| 3,726 | 5,910 | 8,472 |
compute the lower and upper limits (to 2 decimals). If required enter negative value as negative number.
Lower limit____
Upper limit____
Does the data contain any outliers?
first arrange the data in ascending order
| 621 |
| 754 |
| 1383 |
| 1401 |
| 1887 |
| 1910 |
| 2170 |
| 2508 |
| 2874 |
| 3726 |
| 4100 |
| 4428 |
| 5910 |
| 6581 |
| 7628 |
| 8472 |
| 8576 |
| 9057 |
| 10708 |
| 11641 |
| 14138 |
Now, divide the data sets into two equal parts, i.e. bottom 10 data values on one side and top 10 data values on other side
Lower half is
| 621 |
| 754 |
| 1383 |
| 1401 |
| 1887 |
| 1910 |
| 2170 |
| 2508 |
| 2874 |
| 3726 |
First quartile = median of lower half = halfway between two middle values
two middles values are 1887 and 1910
So, First quartile = (1887+1910)/2 = 1898.5 = Q1
Upper half is
| 4428 |
| 5910 |
| 6581 |
| 7628 |
| 8472 |
| 8576 |
| 9057 |
| 10708 |
| 11641 |
| 14138 |
Third quartile = median of upper half = halfway between two middle values
two middles values are 8472 and 8576
So, third quartile = (8472+8576)/2 = 8524 = Q3
Interquartile range or IQR = Q3 - Q1 = 8524 - 1898.5 = 6625.5
Lower limit of data set = Q1- 1.5*IQR
= 1898.5- (1.5*6625.5)
= 1898.5 - 9938.25
= -8039.75
Upper limit of data set = Q3+ 1.5*IQR
= 8524+(1.5*6625.5)
= 8524 + 9938.25
= 18462.25
No, there is no outlier because none of the data set is outside the range (-8039.75, 18462.25)