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Let X1, X2, ..., Xn Be N Observations. Let Y I = a X I + B for I = 1, 2, 3, ..., N, Where a and B Are Constants. - Mathematics

MCQ

Let x1x2, ..., xn be n observations. Let  \[y_i = a x_i + b\]  for i = 1, 2, 3, ..., n, where a and b are constants. If the mean of \[x_i 's\]  is 48 and their standard deviation is 12, the mean of \[y_i 's\]  is 55 and standard deviation of \[y_i 's\]  is 15, the values of a and are 

 
 
 
   

Options

  • a = 1.25, b = −5 

  • a = −1.25, b = 5

  •  a = 2.5, b = −5  

  •  a = 2.5, b = 5 

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Solution

It is given that \[y_i = a x_i + b\]  for i = 1, 2, 3, ..., n, where a and b are constants. 

\[x_i\]  = 48 and \[\sigma_{x_i} = 12\] \[y_i = 55\]  and \[\sigma_{y_i} = 15\]

\[y_i = a x_i + b\]

\[ \Rightarrow \frac{\sum_{} y_i}{n} = \frac{\sum_{} \left( a x_i + b \right)}{n}\]

\[ \Rightarrow \frac{\sum_{} y_i}{n} = a\frac{\sum_{} x_i}{n} + \frac{\sum_{} b}{n}\]

\[ \Rightarrow y_i = a x_i + b \]

\[ \Rightarrow 55 = 48a + b . . . . . \left( 1 \right)\]

Now,
Standard deviation of yi = Standard deviation of \[a x_i + b\]

\[\Rightarrow \sigma_{y_i} = a \times \sigma_{x_i} \]

\[ \Rightarrow 15 = 12a\]

\[ \Rightarrow a = \frac{15}{12} = 1 . 25\]

Putting a = 1.25 in (1), we get
\[b = 55 - 48 \times 1 . 25 = 55 - 60 = - 5\]
Thus, the values of a and b are 1.25 and −5, respectively.
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Q 21 | Page 51
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