Question

In: Math

Finish Men Women 1 66.91 110.64 2 67.88 112.83 3 68.13 113.26 4 68.46 113.54 5...

 
Finish Men Women
1 66.91 110.64
2 67.88 112.83
3 68.13 113.26
4 68.46 113.54
5 72.48 115.99
6 88.79 119.94
7 98.06 122.86
8 100.13 123.69
9 102.13 124.09
10 109.79 124.23
11 110.66 125.49
12 111.84 127.39
13 114.51 131.13
14 115.13 131.48
15 122.56 132.33
16 129.59 133.28
17 130.01 133.64
18 132.51 134.81
19 133.41 135.11
20 140.24 138.18
21 145.44 138.36
22 150.31 139.81
23 140.61
24 148.79
25 148.96
26 149.11
27 149.36
28 155.49
29 156.44
30 190.88
31 190.89

Compare the firstplace finish times for men and women. If the 53 men and women runners had competed as one group, in what place would winner of women's marathon have finished? Round your answer to 2 decimal places.

The first place runner in the men’s group finished  minutes _______aheadbehind of the first place runner in the women’s group.

b. What is the median time for men and women runners? Compare men and women runners based on their median times. Round your answers to 2 decimal places.

Men Women
Median

Using the median finish times, the men’s group finished  minutes _______aheadbehind of the women’s group.

c. Provide a five-number summary for both the men and the women. Round your answers to 2 decimal places.

Men Women
Lowest Time
First Quartile
Median
Third Quartile
Highest Time

d. Are there outliers in either group?

If data contain outliers enter the value. If there no outliers live answer blank. If there is more than one value, separate your answers with commas (to 2 decimals).

Outliers in the men's group.

Outliers in the women's group.

e. Which of the following box plots accurately displays the data set?

#1

Minutes

#2

Minutes

#3

Minutes

#4

Minutes

_________Box plot #1Box plot #2Box plot #3Box plot #4

Did men or women have the most variation in finish times?

_______MenWomen

Solutions

Expert Solution

a). by combining the men runners and women runners upto the the winner women runner we have:-

Finish combined data upto the women winner
1 66.91
2 67.88
3 68.13
4 68.46
5 72.48
6 88.79
7 98.06
8 100.13
9 102.13
10 109.79
11 110.64
12 110.66

the winner of women's marathon would have finished in 11 th place.

The first place runner in the men’s group finished  minutes 43.73 ahead behind of the first place runner in the women’s group.

[ time taken by winner of men's marathon = 66.91 minutes

time taken by winner of women's marathon = 110.64 minutes.

so, the men winner has finished:- (110.64-66.91)= 43.73 minutes before the women winner.]

b). for men,

number of observation (n) =  22

this is an even number.so the median will be:-

for women,

number of observation (n) = 31

this is an odd number.so the median will be:-

so the median of men and women be:-

median of men 111.25
median of women 133.28

c).for men,

first quartile()

[ from the given arranged data of men , 5 th observation = 72.48 and 6 th observation =88.79]

third quartile():-

[ from the given arranged data of men , 17 th observation = 130.01 and 18 th observation =132.51]

FOR WOMEN,

first quartile()

[ from the given arranged data of women , 8 th observation = 123.69 ]

third quartile():-

[ from the given arranged data of women , 24 th observation = 148.79]

so, the five number summary be:-

5 number summary men women
min value 66.91 110.64
first quartile 84.71 123.69
median 111.25 133.28
third quartile 130.63 148.79
max value 150.31 190.89

d).acceptable region= mean 1.5 interquartile range.

for men,

mean = sum of all observation / number of obs

=2378.97/22 = 108.135

IQR = = (130.63-84.71) = 45.92

so, the range be:-

= (108.135 [1.5*45.92] ) = (39.255,177.015)

in MEN group no data is present outside the range, SO NO OUTLIER is present.

for women,

mean = sum of all observation / number of obs

=1212.61 / 31 = 135.89

IQR = = (148.79 - 123.69) = 25.10

so, the range be:-

= (135.89 [1.5*25.10] ) = (98.24 , 173.54)

in WOMEN group two data are present outside the range, SO OUTLIER is present and they are 190.88 , 190.89

e).the boxplot be:-

[steps in minitab:-

graph box plot simple in multiple Y's ok in graph variables select men and women ok.]

men have the most variation in finish times.

[ because the interquartile range of men = 45.92 > interquartile range of women = 25.10

we know that, greater the interquartile range more is the variation. so, MEN has most variation here.]

*** if you have any doubt regarding the problem please write it in the comment box.if you are satisfied please give a LIKE if possible...


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