Question

Use the accompanying data set to complete the following actions.

a. Find the quartiles.
b. Find the interquartile range
c. Identify any outliers.
58 62 63 55 57 62 54 61 56 57 61 56 57 57 76

Answer 1

To find the quartiles, arrange the data in ascending order and find the median, first quartile, and third quartile. The interquartile range is the difference between the third quartile and the first quartile. Outliers can be identified using the 1.5 * IQR rule.

Step-by-step explanation:

To find the quartiles:

1. Arrange the data set in ascending order: 54, 55, 56, 56, 57, 57, 57, 58, 61, 61, 62, 62, 63, 76.
2. Find the median, which is the middle value of the data set: 57.
3. Find the first quartile (Q1), which is the median of the lower half of the data set: (56 + 57) / 2 = 56.5.
4. Find the third quartile (Q3), which is the median of the upper half of the data set: (62 + 62) / 2 = 62.

To find the interquartile range:

The interquartile range is the difference between the third quartile and the first quartile: 62 - 56.5 = 5.5.

To identify any outliers:

An outlier is a data value that is significantly higher or lower than the other values in the data set. To identify outliers, we can use the 1.5 * IQR rule. Any data value less than Q1 - 1.5 * IQR or greater than Q3 + 1.5 * IQR is considered an outlier. In this case, there are no outliers in the data set.

Answer 2

`Q_1=56\\ \\Q_2=57\\ \\Q_3=62\\ \\Q_3-Q_1=9\ \text{Interquartile range}`

There exists at least one outlier in the data set at 76

Step-by-step explanation:

First, arrange the data

58 62 63 55 57 62 54 61 56 57 61 56 57 57 76

in ascending order:

54 55 56 56 57 57 57 57 58 61 61 62 62 63 76

This list of numbers contains 15 numbers, the middle number is 57, then

`Q_2=57`

This number divides the list of numbers in two parts: left part of 7 numbers and right part of 7 numbers.

The middle number of left part is 56, so

`Q_1=56`

The middle number of right part is 62, so

`Q_3=62`

The interquartile range is

`Q_3-Q_1=62-56=6`

Multiply the interquartile range by 1.5:

`1.5(Q_3-Q_1)=1.5\cdot 6=9`

Now subtract it from
`Q_1` and add it to
`Q_3:`

`Q_1-9=56-9=47\\ \\Q_3+9=62+9=71`

The minimum number in the list, 54, is greater than 47, so there is no lower outlier.

The maximum number in the list, 76, is greater than 71, so there is upper outlier.