Mean, Median and Mode Quiz 1

Repeating outputs are
1 - 3 times
6 - 2 times
4 - 2 times

1 is most frequent
So mode = 1

First arrange the outputs in ascending order
1, 1, 1, 2, 3, 4, 4, 5, 6, 6
No. of outputs, n = 10
When n is even, the median is average of \frac{n}{2}th and (\frac{n}{2} + 1)th term.
So here, average of 5rd and 6th terms
Ie, median = \frac{3 + 4}{2}
= 3.5

Reduce 100 from each term to make calculations easy
New series is
8, 5, 11, 7, 9, 6, 4
Sum = 8 + 5 + 11 + 7 + 9 + 6 + 4
= 50
No of observations = 7
mean = \frac{50}{7}
= 7.14

So mean of original series = 100 + 7.14
= 107.14

We have 26 + x + 45 + 13 + 91 = 39 x 5 = 195
x = 195 - 26 - 45 - 13 - 91
x = 20

Now, mean of 15, 32, 48, 12 and x = \frac{15+32+48+12+ x}{5}
=\frac{15+32+48+12+20}{5}
= \frac{127}{5}
=25.4

Sum of 6 numbers = 6 x 10 = 60
Sum of three of them = 3 x 7 = 21
Sum of remaining 3 = 60 - 21 = 39
Mean of remaining 3 numbers = = \frac{39}{3}
= 13

Median is 11. So one of the remaining numbers must be greater than 11 and the other must be less than 11. Middle number will be 11. For maximizing mean, choose the highest possible numbers.
So three numbers = 10, 11, 20

Mean = \frac{10 + 11 + 20}{3}
= \frac{41}{3}
=13.67

Repeating numbers are
7 - 4 times
6 - 2 times
11 - 2 times

7 is the most frequent observation.
So
Mode = 7

Sum of observations = 5 + 15 - 3 + 7 -19 + 5 + 8 - 20 + 2
= 0
Sum of observation is zero
So mean = 0

Sum of observations = 12 x 5 = 60
Median = 16
Sum without considering median observations = 60 - 16 - 16 - 16
= 12
Mean of the other two observations = \frac{12}{2}
= 6

Total goals scored = 1 + 1 + 0 + 3 + 2 + 3 + 1 + 0 + 2 + 4 + 1 + 3
= 21
Total matches played = 12
Mean goals = \frac{21}{12}
= 1.75