Math, asked by needanswer98, 1 month ago

The owner of a coffee shop recorded the number of customers who came into his café each hour in a day. The results were 14, 10, 12, 9, 17, 5, 8, 9, 14, 10, and 11

1. What is the least value of the data?
2. What is the greatest value of the data?
3. Arrange the data in ascending order.
4. Lower Quartile=_____
5. Upper Quartile=_____
6. Interquartile range=_____

Answers

Answered by amitnrw
36

Given :  The owner of a coffee shop recorded the number of customers who came into his café each hour in a day.

The results were 14, 10, 12, 9, 17, 5, 8, 9, 14, 10, and 11

To Find :  

1. What is the least value of the data?

2. What is the greatest value of the data?

3. Arrange the data in ascending order.

4. Lower Quartile=_____

5. Upper Quartile=_____

6. Interquartile range=_____

​Solution :

The results were 14, 10, 12, 9, 17, 5, 8, 9, 14, 10, and 11

Arrange the data in ascending order.

5, 8, 9, 9, 10, 10, 11 , 12,  14, 14,  17  

least value of the data = 5

greatest value of the data = 17

terms  = 11    hence

Lower Quartile=  (11 + 1)/4 = 3 rd term     which is 9

Upper Quartile=  3(11 + 1)/4 = 9th term     which is  14

Interquartile range= Upper Quartile - Lower Quartile

= 14 - 9

= 5

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Answered by TheDiamondBoyy
30

Given:-

The owner of a coffee shop recorded the number of customers who came into his café each hour in a day. The results were 14, 10, 12, 9, 17, 5, 8, 9, 14, 10, and 11.

________________________

Solution:-

The results were 14, 10, 12, 9, 17, 5, 8, 9, 14, 10, and 11

Arrange the data in ascending order.

5, 8, 9, 9, 10, 10, 11 , 12,  14, 14,  17  

→ least value of the data = 5

→ greatest value of the data = 17

→ terms  = 11  

hence,

Lower Quartile =  (11 + 1)/4 = 3 rd term     which is 9

Upper Quartile=  3(11 + 1)/4 = 9th term     which is  14

Interquartile range = Upper Quartile - Lower Quartile

= 14 - 9

= 5

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