Math Olympiad Test: Statistics- 2 - Class 10 MCQ

The given data in ascending order is 25, 31, x - 3, 37, x + 4, 42, 43, 45, 46
n = 9 (odd)

First 5 natural numbers are 1, 2, 3, 4, 5 n = 5 (odd)
∴ Median = n +1/2 th = 5 + 1/2 th = 3rd term
⇒ Median = 3
1, 2, 3, 4, 5, 6
Median = 1/2 (3 + 4) = 7/2 = 3.5
Difference = 3 .5 - 3 = 0.5

N = 8 + 10 + 11 + 16 + 20 + 25 + 15 + 9 + 6 = 120
∴ N/2 = 120/2 = 60
60 lies in between 45 - 65
∴ Variate =  5

Class 12 - 15 has maximum frequency
12 - 15 is the modal class

Given: The mode of a series exceeds its mean by 12.
Formula used: Mode = 3 × Median - 2 × Mean
Calculation: Mode - Mean = 12
As, Mode = 3 × Median - 2 × Mean
⇒ Mode - Mean = 3 × Median - 3 × Mean      [Subtracting Mean from both sides]
⇒ 12 = 3 × (Median - Mean)
⇒ Median - Mean = 12/3 = 4      ----(1)
Again, Mode - Median = 2 × Median - 2 × Mean
⇒ Mode - Median = 2 × 4 = 8
Then, Mode exceeds the median by 8.
∴ The mode exceeds the median by 8.

Here,

Clearly frequency of all 3,4, and 6 are 3. So all of these can be modes of the given distribution.

Concept: Relation between mode, median and mean is given by:
Mode = 3 × median – 2 × mean
Given: Mode – median = 2
As we know
Mode = 3 × median – 2 × mean
Now, Mode = median + 2
⇒ (2 + median) = 3median – 2mean   
⇒ 2Median - 2Mean = 2
⇒ Median - Mean = 1
∴ The difference between the median and mean is 1.

If the no. of observations is even then the median of the data is the average of (n/2) and (n/2 + 1) observations.

Hence median of the data  observations.

The variate having highest frequency is the mode.

The given data in ascending order is 5, 7, 9, 11, 13, 15, 17
n = 7 (odd)

N = 60
N/2 = 30
c.f just greater than 30 is 50
Median class is 30 - 60

    = 30 + 10 = 40

Here the frequency of 8 is maximum. (6 times)
∴ Mode =  8