Question

Let x₁, x₂, ..., x_n 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 b are 

 

 

 

   

पर्याय

• a = 1.25, b = −5 

• a = −1.25, b = 5

•  a = 2.5, b = −5  

•  a = 2.5, b = 5 

Answer 1

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 y_i = 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.