Question

for helping poor girls of their class, student save pocket money as shown in the following table find the median
Money Saved (in rupees): 5-7, 7-9, 9-11, 11-13, 13-15.
No. of students:6,3,9,5,7.
(A)10.33
(B)16.4
(C)9.43​

Answer 1

Solution :-

saved(Rs.) Fi xi di(xi - a/2) fi*di cf

5 - 7 6 6 -2 -12 6

7 - 9 3 8 -1 -3 9

9 - 11 9 10(a) 0 0 18

11 - 13 5 12 1 5 23

13 - 15 7 14 2 14 30

_______________________________

⅀Fi=30. ⅀fi*di=4

_______________________________

Now, we know that,

Median = L + [ (N/2 - F) / f ]H

where :-

• L = Lower boundary of the median class.
• N = Sum of frequencies.
• F = Cumulative frequency before the median class.
• f = Frequency of the median class.
• H = Size of the median class.

So, from our data we have :-

• Median Class = 9 - 11.
• L = 9 .
• N = ⅀Fi= 30 .
• F = 9 .
• f = 9 .
• H = 11 - 9 = 2 .

Putting all values in formula now, we get ,

→ Median = L + [ (N/2 - F) / f ]H

→ Median = 9 + [(30/2 - 9) / 9]2

→ Median = 9 + [(15 - 9) / 9]2

→ Median = 9 + (6/9) * 2

→ Median = 9 + (2/3) * 2

→ Median = 9 + (4/3)

→ Median = 9 + 1 + (1/3)

→ Median = 10 + (1/3)

→ Median = 10.33 (Option A) (Ans.)

Hence, Median of data is 10.33.

Answer 2

Answer:

Step-by-step explanation:

Median class is 9 ? 11.

Median =l+N2−Ff×h

           =9+15−99×2=9+69×2

           =9+1.33=Rs.10.33