Master Question for Statistics

2. Median (Md) 

Md = L + [(N/2 - cfb) / fm] × i

Now, N/2 = 40/2 = 20, and it lies in the class 105-107
Thus, Median Class = 105-107
Md = 104.5 + [(20 - 16) / 9] × 3
= 104.5 + 1.33
= 105.83

3. Mode (Mo) 

First Method - 

Mode = 3 Median - 2 Mean 

Mo = 3 Md - 2 M
Mo = 3 × 105.84 - 2 × 106
= 317.49 - 212 
= 105.49 

Second Method - 

Maximum frequency is 9, thus mode class is 105-107 

Mo = L + [(f1 - f0) / (2 f1 - f0 - f2)] × i
= 104.5 + [(9 - 6) / (18 - 6 - 5)] × 3
= 104.5 + 9/7
= 104.5 + 1.285
= 104.5 + 1.29
= 105.79

4. Range

Range = Upper limit of highest class - Lower limit of lowest class 

Range = 122 - 90
= 32

5. Quartile Deviation 

Q = (Q3 - Q1) / 2 

Q1 = L + [(N/4 - cfb) / f] × i 
= 98.5 + [(10 - 7) / 3] × 3
= 98.5 + 3
= 101.5

Q3 = L + [(3N / 4 - cfb) / f] × i
= 107.5 + [(30 - 25) / 5] × 3
= 107.5 + 3
= 110.5

Q = (110.5 - 101.5) / 2
= 9/2
= 4.5