Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 40,33,37,54,41,34,27,39,35 Item 4 Find the median, first quartile, third quartile, and interquartile range of the data. 40,33,37,54,41,34,27,39,35

Median =37

First quartile = 33.5

Third quartile=40.5

Interquartile=7

Step-by-step explanation:

Given data

40,33,37,54,41,34,27,39,35

First we have to rearrange the data in either ascending order or descending order.

We can rewrite the data as

27,33,34,35,37,39,40,41,54

Median:

The middle term of the data is the median of the data.

The number of data = n

If n is odd, the median of the data= term

If n is even, the median of the data

Here n= 9

The median of the data= term

                                        term

                                        term

                                       =37

First quartile:

The median term of the lower half of the data.

Lower half = 27,33,34,35

The median of the lower half =

                                               =

                                              =

                                               =33.5

The first quartile= 33.5

Third  quartile:

The median term of the upper half of the data.

Upper half 39,40,41,54

The median of the lower half =

                                                =

                                                =

                                                 =40.5

Interquartile:

The difference between first quartile and third quartile.

Interquartile = 40.5-33.5

                       =7

3030 ur welcome

step-by-step explanation:

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40