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Principle of Mathematical Induction
notes
By using step-deviation method, it is possible to simplify the procedure.
Let the assumed mean be ‘A’ and the scale be reduced to `1/h` times (h being the width of class-intervals). Let the step-deviations or the new
values be `y_i`.
i.e. `y_i = (x_i - A) /h or x_i = A +hy_i` ...(1)
we know that \[\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} f_ix_i}{N} \]
Replacing xi from (1) in (2), we get
\[\bar{x} =\frac{\displaystyle\sum_{i=1}^{n} f_i(A + hy_i)}{N} \]
=\[\frac {1}{N} (\displaystyle\sum_{i=1}^{n}
f_i A + = \displaystyle\sum_{i=1}^{n}
hf_iy_i )\] = =\[\frac {1}{N} (A \displaystyle\sum_{i=1}^{n}
f_i + h = \displaystyle\sum_{i=1}^{n}
f_i y_i) \]
=A .\[\frac{N}{N} + h \frac {\displaystyle\sum_{i=1}^{n} f_iy_i}{N} \]
(because \[\displaystyle\sum_{i=1}^{n} f_i\] = N )
Thus `bar x = A + h bar y` ...(3)
Now Variance of the variable x, \[\sigma_x^2 =\frac {1}{N} \displaystyle\sum_{i=1}^{n}
f_i (x_i - \bar x)^2 \]
=\[\frac {1}{N} \displaystyle\sum_{i=1}^{n}
f_i (A +hy_i - A - h \bar y)^2 \] (Using (1) and (2))
=\[\frac {1}{N} \displaystyle\sum_{i=1}^{n}
f_i h^2 (y_i - \bar y)^2 \]
= \[\frac {h^2}{N} \displaystyle\sum_{i=1}^{n}
f_i (y_i -\bar y)^2 = h^2 × \] variance of the variable yi
i .e.`sigma_x^2 = h^2 sigma_x^2`
or `σ _x = hσ _y` ...(4)
From (3) and (4), we have
σx = \[{\frac{h}{N}}\sqrt{N\displaystyle\sum_{i=1}^{n} f_i y_i ^2 - (\displaystyle\sum_{i=1}^{n} f_i y_i) ^2 } \] ...(5)
Shaalaa.com | Shortcut method for Standard Deviation
Related QuestionsVIEW ALL [9]
There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test.
Marks | 0 | 1 | 2 | 3 | 4 | 5 |
Frequency | x – 2 | x | x2 | (x + 1)2 | 2x | x + 1 |
where x is a positive integer. Determine the mean and standard deviation of the marks.
Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
Ravi | 25 | 50 | 45 | 30 | 70 | 42 | 36 | 48 | 35 | 60 |
Hashina | 10 | 70 | 50 | 20 | 95 | 55 | 42 | 60 | 48 | 80 |
Who is more intelligent and who is more consistent?
The frequency distribution:
`x` | A | 2A | 3A | 4A | 5A | 6A |
`f` | 2 | 1 | 1 | 1 | 1 | 1 |
where A is a positive integer, has a variance of 160. Determine the value of A.
The weights of coffee in 70 jars is shown in the following table:
Weight (in grams) |
Frequency |
200 – 201 | 13 |
201 – 202 | 27 |
202 – 203 | 18 |
203 – 204 | 10 |
204 – 205 | 1 |
205 – 206 | 1 |
Determine variance and standard deviation of the above distribution.
From the prices of shares X and Y below, find out which is more stable in value:
X |
35 |
54 |
52 |
53 |
56 |
58 |
52 |
50 |
51 |
49 |
Y |
108 |
107 |
105 |
105 |
106 |
107 |
104 |
103 |
104 |
101 |
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results::
Firm A |
Firm B |
|
No. of wage earners |
586 |
648 |
Mean of monthly wages |
Rs 5253 |
Rs 5253 |
Variance of the distribution of wages |
100 |
121 |
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
Find the mean and standard deviation using short-cut method.
xi |
60 |
61 |
62 |
63 |
64 |
65 |
66 |
67 |
68 |
fi |
2 |
1 |
12 |
29 |
25 |
12 |
10 |
4 |
5 |