मराठी

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 Median - Mathematics

Advertisements
Advertisements

प्रश्न

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 median

थोडक्यात उत्तर

उत्तर

 Formula for the mean deviation from the median is as follows:

\[MD = \frac{1}{n} \sum^n_{i = 1} \left| d_i \right|, \text{ where } \left| d_i \right| = \left| x_i - M \right|\]
Arranging the data in ascending order for finding the median:
15.2, 27.9, 30.2, 32.5, 40, 52.3, 52.8, 55.2, 72.9, 79
Here, 
n = 10. 
Therefore, median is the average of the fifth and the sixth observations.
\[M = \frac{40 + 52 . 3}{2} = 46 . 15\]
\[x_i\]
\[\left| d_i \right| = \left| x_i - 46 . 15 \right|\]
40 6.15
52.3 6.15
55.2 9.05
72.9 26.75
52.8 6.65
79 32.85
32.5 13.65
15.2 30.95
27.9 18.25
30.2 15.95
Total 166.4

\[MD = \frac{1}{10} \times 166.4 = 16.64\]

Mean deviation from median in 16.64cm.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 32: Statistics - Exercise 32.1 [पृष्ठ ६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 32 Statistics
Exercise 32.1 | Q 4.1 | पृष्ठ ६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the mean deviation about the median for the data.

36, 72, 46, 42, 60, 45, 53, 46, 51, 49


Find the mean deviation about the mean for the data.

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

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.

Income per day in ₹ Number of persons
0-100 4
100-200 8
200-300 9
300-400 10
400-500 7
500-600 5
600-700 4
700-800 3

Calculate the mean deviation about the median of the observation:

 38, 70, 48, 34, 42, 55, 63, 46, 54, 44


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: 

 4, 7, 8, 9, 10, 12, 13, 17


Calculate the mean deviation from the mean for the  data:

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


Calculate the mean deviation of the following income groups of five and seven members from their medians:

I
Income in Rs.
II
Income in Rs.
4000
4200
4400
4600
4800

 
 300
4000
4200
4400
4600
4800
5800

In 38, 70, 48, 34, 63, 42, 55, 44, 53, 47 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 10 30 50 70 90
fi 4 24 28 16 8

Find the mean deviation from the median for the  data:

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

 


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

 


Find the mean deviation from the mean for the data:

Classes 0-10 10-20 20-30 30-40 40-50 50-60
Frequencies 6 8 14 16 4 2

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

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.

 

The mean deviation of the numbers 3, 4, 5, 6, 7 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 distribution:

Size 20 21 22 23 24
Frequency 6 4 5 1 4

Find the mean deviation about the median of the following distribution:

Marks obtained 10 11 12 14 15
No. of students 2 3 8 3 4

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.


Calculate the mean deviation about the mean for the following frequency distribution:

Class interval 0 – 4 4 – 8 8 – 12 12 – 16 16 – 20
Frequency 4 6 8 5 2

Calculate the 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

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.


Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.


The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is ______.


If `barx` is the mean of n values of x, then `sum_(i = 1)^n (x_i - barx)` is always equal to ______. If a has any value other than `barx`, then `sum_(i = 1)^n (x_i - barx)^2` is ______ than `sum(x_i - a)^2`


The sum of squares of the deviations of the values of the variable is ______ when taken about their arithmetic mean.


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 ______.


Find the mean deviation about the mean for the data.

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×