A. Calculate the median and interquartile range for each set of data. Make sure to label each measurement in your answer. B. Make two inferences based on your results from part A. Use complete sentences in your answer

kindly check Explanation

Step-by-step explanation:

1948 Olympics: Men's 100m Freestyle swim finals

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6, 65

Lower quartile (Q1) : 1/4(n+1)th term

Where n = number of observations

Q1 = 1/4(8+1)th term

Q1 = 9/4 = 2.25 term

Average of 2nd and 3rd term:

(57.8 + 58.1) / 2 = 115.9/2 = 57.95

Median (Q2) :

Q2 = 1/2(n+1)th term

Q2 = 1/2(9)th term

Q2 = 9/2 = 4.5th term

Average of 4th and 5th term:

(58.3 + 58.3) / 2 = 58.3

Upper quartile (Q3) :

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

Where n = number of observations

Q3 = 3/4(8+1)th term

Q3 = 27/4 = 6.75 term

Average OF 6th and 7th term:

(59.3 + 59.6) / 2 = 118.9/2 = 59.45

Interquartile range ( IQR) :

(Q3 - Q1)

59.45 - 57.95 = 1.5

2) 2012 Olympics: Men's 100m Freestyle swim finals

47.52, 47.53, 47.80, 47.84, 47.88, 47.92, 48.04, 48.44

Lower quartile (Q1) : 1/4(n+1)th term

Where n = number of observations

Q1 = 1/4(8+1)th term

Q1 = 9/4 = 2.25 term

Average of 2nd and 3rd term:

(47.53+ 47.80) / 2 = 95.33/2 = 47.67

Median (Q2) :

Q2 = 1/2(n+1)th term

Q2 = 1/2(9)th term

Q2 = 9/2 = 4.5th term

Average of 4th and 5th term:

(47.84 + 47.88) / 2 = 47.86

Upper quartile (Q3) :

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

Where n = number of observations

Q3 = 3/4(8+1)th term

Q3 = 27/4 = 6.75 term

Average OF 6th and 7th term:

(47.92 + 48.04) / 2 = 47.98

Interquartile range ( IQR) :

(Q3 - Q1)

47.98 - 47.67 = 0.31

Range of values in the 2012 Olympics Swim finals are closer to one another than in the 1948 Olympics as shown by the values of the interquartile range.

The athlete's Swim speed seems to have improved drastically from the 1948 Olympics during the 2012 finals as athletes elapsed time have become significantly lower.

answer: approximately 206 bones

i hope this u out a branlist and a would be greatly appreciated :