# 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

#### 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?

#### APPEARS IN

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