Question

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. The box plot compares the monthly average temperature (in degrees Fahrenheit) recorded in the towns of Springwood and Meadows from April to October. Match each phrase to its correct value.
the median of the temperatures at Springwood _______


the median of the temperatures at Meadows: ______


the interquartile range of the temperatures at Springwood:_______


the interquartile range of the temperatures at Meadows:_______


the difference of the medians as a multiple of their average interquartile range:_______

Answer 1

Answer: The median of the temperatures at Springwood is 86.    

The median of the temperatures at Meadows is 73.  

The interquartile range of the temperatures at Meadows is 12.  

The interquartile range of the temperatures at Springwood is 14.

The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13

i.e. Difference in medians  = 1 x (Average interquartile range)

Step-by-step explanation:

In a box - whisker plot , the vertical line in box represents the median value.

The left end of box denotes Lower quartile and the right end of the box Upper quartile.

By considering the given picture,

For Meadows ,

Median = 73

The median of the temperatures at Meadows is 73.                           (1)

Lower quartile = 68

Upper quartile = 80

Interquartile range = Upper quartile- Lower quartile

Interquartile range=80-68 = 12

The interquartile range of the temperatures at Meadows is 12.     (2)

For Springwood,

Median = 86

The median of the temperatures at Springwood is 86.                 (3)

Lower quartile = 77

Upper quartile = 91

Interquartile range = Upper quartile- Lower quartile

= 91-77 = 14

The interquartile range of the temperatures at Springwood is 14.         (4)

Difference in medians = 86-73 = 13              [From (1) and (3)]

Average interquartile range = $\dfrac{12+14}{2}=13$    [From (2) and (4)]

The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13

i.e. Difference in medians = 1 x (Average interquartile range)

Answer 2

Hello,

The median of the temperatures at Springwood is 86. (median is at the middle).

the median of the temperatures at Meadows is 73. (median is at the middle).

The (IQR) interquartile range of the temperatures at Meadows is 12. (You have to subtract both of the IQR left and right 80 - 68 = 12).

The (IQR) interquartile range of the temperatures at Springwood is 14. (You have to subtract both of the IQR left and right 91 - 77 = 14).

the difference of the medians as a multiple of their average interquartile range is 79.5 is average median and 13 average IQR.