Question

consider the time taken to completion time (in months) for the construction of a particular model of homes: 4.1 3.2 2.8 2.6 3.7 3.1 9.4 2.5 3.5 3.8 Find the mean, median mode, first quartile and third quartile. Find the outlier?​

Answer 1

Sum

Mean = (4.1 + 3.2 + 2.8 + 2.6 + 3.7 + 3.1 + 9.4 + 2.5 + 3.5 + 3.8) / 10

Mean = 36.7 / 10

Mean = 3.67

To find the median, we need to put the values in order:

2.5, 2.6, 2.8, 3.1, 3.2, 3.5, 3.7, 3.8, 4.1, 9.4

The middle number is the median, which is 3.35 in this case (3.2+3.5)/2.

To find the mode, we look for the value that appears most often. In this case, there is no mode as no value appears more than once.

To find the first quartile (Q1), we need to find the value that separates the bottom 25% of the data from the top 75%. We can do this by finding the median of the lower half of the data:

2.5, 2.6, 2.8, 3.1, 3.2

The median of this lower half is 2.8, so Q1 = 2.8.

To find the third quartile (Q3), we need to find the value that separates the bottom 75% of the data from the top 25%. We can do this by finding the median of the upper half of the data:

3.7, 3.8, 4.1, 9.4

The median of this upper half is 3.95, so Q3 = 3.95.

To find the outlier, we can use the rule that any value more than 1.5 times the interquartile range (IQR) away from the nearest quartile is considered an outlier. The IQR is the difference between Q3 and Q1:

IQR = Q3 - Q1

IQR = 3.95 - 2.8

IQR = 1.15

1.5 times the IQR is 1.5 * 1.15 = 1.725.

The only value that is more than 1.725 away from either Q1 or Q3 is 9.4. Therefore, 9.4 is the outlier in this data set.

Answer 2

Explanation:

Mean is the average of the data vales : 38.7 (sum of all values) divided by 10 (the number of values). Mean = 38.7/10 = 3.87

Median is the "middle" number" = put the date in order and find the middle value:

2.5, 2.6, 2.8, 3.1, 3.2, 3.5, 3.7, 3.8, 4.1, 9.4, since there is no middle data value, find the average of the 2 in the middle.

3.2 + 3.5/2 = 6.7 ÷ 2 = 3.35 - median

All values appear once so there is no mode.

First quartile is the middle of the lower set of data = 2.8

Third quartile is the middle of the upper set = 3.8

The outlier is 9.4.

How to calculate the outlier:

First you need the IQR which is the diffence of Q3 - Q1,

so the IQR is 3.8- 2.8 = 1

Outliers are the quartile + or - (1.5)(IQR)

Q1 -(1.5)(1) = 2.8 - 1.5 = 1.3

Q3 - (1.5)(1)= 3.8 + 1.5 =4.5

So anything below 1.3 or above 4.5 is an outlier.

There is one 9.4