मराठी

Calculate Mean Deviation from the Median of the Following Data: Class Interval: 0–6 6–12 12–18 18–24 24–30 Frequency 4 5 3 6 2 - Mathematics

Advertisements
Advertisements

प्रश्न

Calculate mean deviation from the median of the following data: 

Class interval: 0–6 6–12 12–18 18–24 24–30
Frequency 4 5 3 6 2
Advertisements

उत्तर

Calculation of mean deviation about the median. 

Class Interval Mid-Values
(xi)
Frequency
(fi)
Cummulative
Frequency (c.f.)
 

\[\left| x_i - 14 \right|\]
 

\[f_i \left| x_i - 14 \right|\]
0–6 3 4 4 11 44
6–12 9 5 9 5 25
12–18 15 3 12 1 3
18–24 21 6 18 7 42
24–30 27 2 20 13 26
    N = 20    
 

\[\sum f_i| x_i - 14 | = 140\]

Here, N = 20. So,

\[\frac{N}{2} = 10\] The cummulative frequency just greater than \[\frac{N}{2}\]  is 12. Thus, 12–18 is the median class.
 
Now, l = 12, h = 6, f = 3 and F = 9
\[\therefore \text{ Median } = l + \frac{\frac{N}{2} - F}{f} \times h = 12 + \left( \frac{10 - 9}{3} \right) \times 6 = 14\]
Now,

Mean deviation about median = \[\frac{1}{N}$\sum_{} f_i \left| x_i - 14 \right| = \frac{1}{20} \times 140 = 7\]
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 32: Statistics - Exercise 32.3 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 32 Statistics
Exercise 32.3 | Q 8 | पृष्ठ १७

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the mean deviation about the mean for the data.

38, 70, 48, 40, 42, 55, 63, 46, 54, 44


Find the mean deviation about the median for the data.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17


Find the mean deviation about the median for the data.

xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

Find the mean deviation about the median for the data.

xi 15 21 27 30 35
fi 3 5 6 7 8

Find the mean deviation about the mean for the data.

Height in cms Number of boys
95 - 105 9
105 - 115 13
115 - 125 26
125 - 135 30
135 - 145 12
145 - 155 10

Calculate the mean deviation about the median of the observation:

3011, 2780, 3020, 2354, 3541, 4150, 5000


Calculate the mean deviation about the median of the observation:

 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

 

Calculate the mean deviation from the mean for the  data:

 38, 70, 48, 40, 42, 55, 63, 46, 54, 44a


The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2 

Find mean deviation from the mean also.

 

 


In 34, 66, 30, 38, 44, 50, 40, 60, 42, 51 find the number of observations lying between

\[\bar{ X } \]  − M.D. and

\[\bar{ X } \]  + M.D, where M.D. is the mean deviation from the mean.


In  22, 24, 30, 27, 29, 31, 25, 28, 41, 42 find the number of observations lying between 

\[\bar { X } \]  − M.D. and

\[\bar { X } \]   + M.D, where M.D. is the mean deviation from the mean.


Find the mean deviation from the mean for the data:

xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

Find the mean deviation from the mean for the data:

xi 5 10 15 20 25
fi 7 4 6 3 5

Find the mean deviation from the mean for the data:

xi 10 30 50 70 90
fi 4 24 28 16 8

Find the mean deviation from the mean for the data:

Size 1 3 5 7 9 11 13 15
Frequency 3 3 4 14 7 4 3 4

Find the mean deviation from the median for the data: 

xi 74 89 42 54 91 94 35
fi 20 12 2 4 5 3 4

Find the mean deviation from the mean for the data:

Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Frequencies 4 8 9 10 7 5 4 3

 


Find the mean deviation from the mean for the data:

Classes 95-105 105-115 115-125 125-135 135-145 145-155
Frequencies 9 13 16 26 30 12

 


Compute mean deviation from mean of the following distribution:

Mark 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of students 8 10 15 25 20 18 9 5

The age distribution of 100 life-insurance policy holders is as follows:

Age (on nearest birth day) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5
No. of persons 5 16 12 26 14 12 6 5

Calculate the mean deviation from the median age


Find the mean deviation from the mean and from median of the following distribution:

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 5 8 15 16 6

Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age: 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55
Number of persons 5 6 12 14 26 12 16 9

The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

 

A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is


The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is


The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is


The mean deviation for n observations \[x_1 , x_2 , . . . , x_n\]  from their mean \[\bar{X} \]  is given by

 
  

Find the mean deviation about the mean of the following data:

Size (x): 1 3 5 7 9 11 13 15
Frequency (f): 3 3 4 14 7 4 3 4

The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.


Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.


Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.


Let X = {x ∈ N: 1 ≤ x ≤ 17} and Y = {ax + b: x ∈ X and a, b ∈ R, a > 0}. If mean and variance of elements of Y are 17 and 216 respectively then a + b is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×