Question

11.Find the mean deviation about median for the following data:MarksNumber ofGirls0-10 10-20 20-30 30-40 40-50 50-6014164

Answer 1

The given data is:

Marks	Number of girls
0-10	6
10-20	8
20-30	14
30-40	16
40-50	4
50-60	2

The midpoint of the intervals is calculated by adding the first and last terms and dividing it by 2. The cumulative frequencies is given by writing the first term of discrete frequency in the first row of cumulative frequency then adding the corresponding terms of cumulative frequency and frequency. The formula to calculate the absolute deviation of the respective observations of the data is,

| x i −M | .

Now, make a table using the above formula and continuous frequency.

Marks	Number of girls, f i	c.f.	Midpoint, x i	| x i −M |	f i | x i −M |
0-10	6	6	5	22.85	137.1
10-20	8	6+8=14	15	12.85	102.8
20-30	14	14+14=28	25	2.85	39.9
30-40	16	28+16=44	35	7.15	114.4
40-50	4	44+4=48	45	17.15	68.6
50-60	2	48+2=50	55	27.15	54.3
	∑ i=1 n f i =50				∑ i=1 n f i | x i − x ¯ | =517.1

Identify the observation whose cumulative frequency is equal to or just greater than N 2 , where N is the sum of frequency.

Since N = 50, therefore the 25th term lies in the class 20-30 with c.f. 28. Therefore, the median class is 20-30.

The formula to calculate the median is,

M=l+ N 2 −C f ×h

Here,l is the lower limit of the median class, C is the cumulative frequency of the class preceding the median class, fis the frequency corresponding to the median class and h is the interval gap (Class size).

M=20+ 50 2 −14 14 ×10 =20+ 25−14 14 ×10 =20+ 110 14 =27.85

Therefore, the median of the given data is 27.85.

The formula to calculate the mean deviation about the median is,

M.D= ∑ i=1 n f i | x i −M | ∑ i=1 n f i M.D= 1 N ∑ i=1 n f i | x i −M |

M.D.= 517.1 50 =10.34

Therefore, the mean deviation of the given data is 10.34.