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The Standard Deviation of the Data:X:1aa2....Anf:Nc0nc1nc2....Ncn is

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प्रश्न

The standard deviation of the data:

x: 1 a a2 .... an
f: nC0 nC1 nC2 .... nCn

is

विकल्प

  • \[\left( \frac{1 + a^2}{2} \right)^n - \left( \frac{1 + a}{2} \right)^n\]

     
  •  \[\left( \frac{1 + a^2}{2} \right)^{2n} - \left( \frac{1 + a}{2} \right)^n\]

  •  \[\left( \frac{1 + a}{2} \right)^{2n} - \left( \frac{1 + a^2}{2} \right)^n\]

     

  • none of these

     
MCQ
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उत्तर

none of these

xi fi fixi
 

\[{x_i}^2\]
 

\[f_i {x_i}^2\]
1
 

\[^{n}{}{C}_0\]
\[^{n}{}{C}_0\]
1 1
a
\[{n}{}{C}_1\]
\[^{n}{}{C}_1\]
a2 a2 
\[^{n}{}{C}_1\]
a2
\[^{n}{}{C}_2\]
a2
=\[^{n}{}{C}_2\]
a4   a4 
\[^{n}{}{C}_2\]
a3
 

\[^{n}{}{C}_3\]
 a3
\[^{n}{}{C}_3\]
a6  a6  
 

\[^{n}{}{C}_3\]
:
:
:
:
:
:
:
:
:
:
:
:
:
 
:
:
:
:
an
 

\[^{n}{}{C}_n\]
an 
\[^{n}{}{C}_n\]
a2n a2n
\[^{n}{}{C}_n\]
 
\[\sum^n_{i = 1} f_i = 2^n\]
\[\sum^n_{i = 1} f_i x_i = \left( 1 + a \right)^n\]
 
\[\sum^n_{i = 1} f_i {x_i}^2 = \left( 1 + a^2 \right) {}^n\]

`"Number of terms," N = \sum_{i = 1}^2  f_i = 2^n `

` \sum _{i = 1}^2 f_i x_i = ^nC_0 + a ^nC_1 + a^2 "^nC_2 + . . . + a"^n "^nC_n = \left( 1 + a \right)^n `

`X = \frac{\sum_{i = 1}^n f_ix_i}{N}`

\[ = \frac{\left( 1 + a \right)^n}{2^n}\]

` \sum_{i = 1}^n f_i x_i^2 = \left( 1 + a^2 \right)^n`

`\sigma^2 = \text{ Variance } \left( X \right) = \frac{1}{N} \sum_{i = 1}^n f_i_x_i^2 - \left( {\sum_{i = 1}^n f_i x_i}/{N} \right)^2 `

\[ = \frac{\left( 1 + a^2 \right)^n}{2^n} - \left[ \frac{\left( 1 + a \right)^n}{2^n} \right]^2 \]

\[ = \left[ \frac{1 + a^2}{2} \right]^n - \left[ \frac{1 + a}{2} \right]^{2n} \]

\[\sigma = \sqrt{\text{ Variance }  \left( X \right)} \]

\[ = \sqrt[]{\left[ \frac{1 + a^2}{2} \right]^n - \left[ \frac{1 + a}{2} \right]^{2n}}\]

 
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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 32: Statistics - Exercise 32.9 [पृष्ठ ५०]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 32 Statistics
Exercise 32.9 | Q 6 | पृष्ठ ५०

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