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masha68 [24]
3 years ago
6

The median of Rae's data is 2.22 minutes, and the median of Doris' data is 2.14 minutes. The interquartile range for Rae is , an

d the interquartile range for Doris is , . The two data sets overlap
Mathematics
1 answer:
JulsSmile [24]3 years ago
6 0

Complete question :

Rae and Doris are training to swim a 200-meter freestyle race. The table lists their practice times during training camp.

Rae’s Times(minutes)-2.12, 2.01, 2.46, 2, 2.22, 2.31, 2.23

doris times(minutes)-2.32, 2.19, 2.26, 2.03, 2.11, 2.14, 2.07

The median of Raes data is(drop down box answers, 2,2.03,2.14,2.22)minutes,and the median of doris data is(drop down box answers, 2,2.03,2.14,2.22)minutes. The interquartile range for rae is(drop down box answers, 0.01,0.13,0.26,0.30), and the interquartile range for doris is(drop down box answers, 0.02,0.06,0.18,0.19). The two data sets overlap(drop down box answers,very little, a lot

Answer:

Rae's median = 2.22

Rae's IQR = 0.30

DORIS median = 2.14

Doris IQR = 0.19

Step-by-step explanation:

Given :

Rae's time :

Ordered data : 2, 2.01, 2.12, 2.22, 2.23, 2.31, 2.46

Median = 1/2(n+1)th term ; n = 7

Median = 1/2(8) = 4th term

Median = 2.22

Q1 = 1/4(8)th term

Q1 = 2nd term = 2.01

Q3 = 3/4(n + 1)th term

Q3 = 3/4(8)th term

Q3 = 6th term = 2.31

Interquartile range = Q3 - Q1 = 2.31 - 2.01 = 0.30

DORIS Data:

Ordered data: 2.03, 2.07, 2.11, 2.14, 2.19, 2.26, 2.32

Median = 1/2(n+1)th term ; n = 7

Median = 1/2(8) = 4th term

Median = 2.14

Q1 = 1/4(8)th term

Q1 = 2nd term = 2.07

Q3 = 3/4(n + 1)th term

Q3 = 3/4(8)th term

Q3 = 6th term = 2.26

Interquartile range = Q3 - Q1 = 2.26 - 2.07 = 0.19

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