Question 1. Calculate the mean deviation from the median of the following frequency distribution:
Heights in inches |
58 |
59 |
60 |
61 |
62 |
63 |
64 |
65 |
66 |
No. Of Students |
15 |
20 |
32 |
35 |
35 |
22 |
20 |
10 |
8 |
Solution:
Median is the middle term of the observation in ascending order,
So, Median = 61
Let us assume,
xi =Heights in inches
fi = Number of students
xi |
fi |
Cumulative Frequency |
|di| = |xi – M|
= |xi – 61|
|
fi |di| |
58 |
15 |
15 |
3 |
45 |
59 |
20 |
35 |
2 |
40 |
60 |
32 |
67 |
1 |
32 |
61 |
35 |
102 |
0 |
0 |
62 |
35 |
137 |
1 |
35 |
63 |
22 |
159 |
2 |
44 |
64 |
20 |
179 |
3 |
60 |
65 |
10 |
189 |
4 |
40 |
66 |
8 |
197 |
5 |
40 |
|
N = 197 |
|
|
Total = 336 |
N=197

= 1/197 × 336
= 1.70
Therefore, the mean deviation is 1.70.
Question 2. The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:
Number of calls |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
Frequency |
14 |
21 |
25 |
43 |
51 |
40 |
39 |
12 |
Compute the mean deviation about the median.
Solution:
Median is the middle term of the observation in ascending order,
We know, Median is the even term, (3+5)/2 = 4
So, Median = 8
Let us assume
xi =Number of calls
fi = Frequency
N = 245
xi |
fi |
Cumulative Frequency |
|di| = |xi – M|
= |xi – 61|
|
fi |di| |
0 |
14 |
14 |
4 |
56 |
1 |
21 |
35 |
3 |
63 |
2 |
25 |
60 |
2 |
50 |
3 |
43 |
103 |
1 |
43 |
4 |
51 |
154 |
0 |
0 |
5 |
40 |
194 |
1 |
40 |
6 |
39 |
233 |
2 |
78 |
7 |
12 |
245 |
3 |
36 |
|
Total = 245 |
|
|
Total = 366 |

= 1/245 × 336
= 1.49
Therefore, mean deviation is 1.49.
Question 3. Calculate the mean deviation about the median of the following frequency distribution:
xi |
5 |
7 |
9 |
11 |
13 |
15 |
17 |
fi |
2 |
4 |
6 |
8 |
10 |
12 |
8 |
Solution:
Calculating the median,
We know, Number of observations, N = 50
Median = (50)/2 = 25
Therefore, the median corresponding to 25 is 13.
xi |
fi |
Cumulative Frequency |
|di| = |xi – M|
= |xi – 61|
|
fi |di| |
5 |
2 |
2 |
8 |
16 |
7 |
4 |
6 |
6 |
24 |
9 |
6 |
12 |
4 |
24 |
11 |
8 |
20 |
2 |
16 |
13 |
10 |
30 |
0 |
0 |
15 |
12 |
42 |
2 |
24 |
17 |
8 |
50 |
4 |
32 |
|
Total = 50 |
|
|
Total = 136 |

= 1/50 × 136
= 2.72
Therefore, the mean deviation is 2.72.
Question 4. Find the mean deviation from the mean for the following data:
(i)
xi |
5 |
7 |
9 |
10 |
12 |
15 |
fi |
8 |
6 |
2 |
2 |
2 |
6 |
(ii)
xi |
5 |
10 |
15 |
20 |
25 |
fi |
7 |
4 |
6 |
3 |
5 |
(iii)
xi |
10 |
30 |
50 |
70 |
90 |
fi |
4 |
24 |
28 |
16 |
8 |
Solution:
(i) We know,

xi |
fi |
Cumulative Frequency (xifi) |
|di| = |xi – Mean| |
fi |di| |
5 |
8 |
40 |
4 |
32 |
7 |
6 |
42 |
2 |
12 |
9 |
2 |
18 |
0 |
0 |
10 |
2 |
20 |
1 |
2 |
12 |
2 |
24 |
3 |
6 |
15 |
6 |
90 |
6 |
36 |
|
Total = 26 |
Total = 234 |
|
Total = 88 |
Now, Mean = 234/26
= 9

= 88/26
= 3.3
∴ The mean deviation is 3.3
(ii) We know,

xi |
fi |
Cumulative Frequency (xifi) |
|di| = |xi – Mean| |
fi |di| |
5 |
7 |
35 |
9 |
63 |
10 |
4 |
40 |
4 |
16 |
15 |
6 |
90 |
1 |
6 |
20 |
3 |
60 |
6 |
18 |
25 |
5 |
125 |
11 |
55 |
|
Total = 25 |
Total = 350 |
|
Total = 158 |
Mean = 350/25
= 14

= 158/25
= 6.32
∴ The mean deviation is 6.32
(iii) We know,

= 4000/80
= 50
xi |
fi |
Cumulative Frequency (xifi) |
|di| = |xi – Mean| |
fi |di| |
10 |
4 |
40 |
40 |
160 |
30 |
24 |
720 |
20 |
480 |
50 |
28 |
1400 |
0 |
0 |
70 |
16 |
1120 |
20 |
320 |
90 |
8 |
720 |
40 |
320 |
|
Total = 80 |
Total = 4000 |
|
Total = 1280 |

= 1280/80
= 16
∴ The mean deviation is 16
Question 5. Find the mean deviation from the median for the following data :
(i)
xi |
15 |
21 |
27 |
30 |
fi |
3 |
5 |
6 |
7 |
(ii)
xi |
74 |
89 |
42 |
54 |
91 |
94 |
35 |
fi |
20 |
12 |
2 |
4 |
5 |
3 |
4 |
(iii)
Marks obtained |
10 |
11 |
12 |
14 |
15 |
No. of students |
2 |
3 |
8 |
3 |
4 |
Solution:
(i) We know,
Number of observations, N = 21
Median = (21)/2 = 10.5
Therefore, the median corresponding to 10.5 is 27
xi |
fi |
Cumulative Frequency |
|di| = |xi – M| |
fi |di| |
15 |
3 |
3 |
15 |
45 |
21 |
5 |
8 |
9 |
45 |
27 |
6 |
14 |
3 |
18 |
30 |
7 |
21 |
0 |
0 |
|
Total = 21 |
Total = 46 |
|
Total = 108 |

= 1/21 × 108
= 5.14
∴ The mean deviation is 5.14
(ii) We know,
Number of observations, N =50
Median = (50)/2 = 25
Therefore, the median corresponding to 25 is 74.
xi |
fi |
Cumulative Frequency |
|di| = |xi – M| |
fi |di| |
74 |
20 |
4 |
39 |
156 |
89 |
12 |
6 |
32 |
64 |
42 |
2 |
10 |
20 |
80 |
54 |
4 |
30 |
0 |
0 |
91 |
5 |
42 |
15 |
180 |
94 |
3 |
47 |
17 |
85 |
35 |
4 |
50 |
20 |
60 |
|
Total = 50 |
Total = 189 |
|
Total = 625 |

= 1/50 × 625
= 12.5
Therefore, the mean deviation is 12.5
(iii) We know,
Number of observations, N =20
Median = (20)/2 = 10
So, the median corresponding to 10 is 12.
xi |
fi |
Cumulative Frequency |
|di| = |xi – M| |
fi |di| |
10 |
2 |
2 |
2 |
4 |
11 |
3 |
5 |
1 |
3 |
12 |
8 |
13 |
0 |
0 |
14 |
3 |
16 |
2 |
6 |
15 |
4 |
20 |
3 |
12 |
|
Total = 20 |
|
|
Total = 25 |

= 1/20 × 25
= 1.25
∴ The mean deviation is 1.25
Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape,
GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out -
check it out now!
Last Updated :
11 Feb, 2021
Like Article
Save Article