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 $x_i\text{'}\mathrm{s}$ is 48 and their standard deviation is 12, the mean of $y_i\text{'}\mathrm{s}$ is 55 and standard deviation of $y_i\text{'}\mathrm{s}$ is 15, the values of a and b are

(a) a = 1.25, b = −5 (b) a = −1.25, b = 5 (c) a = 2.5, b = −5 (d) 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.

$\overline{x_i}$ = 48 and ${\sigma }_{x_i}=12$

$\overline{y_i}=55$ and ${\sigma }_{y_i}=15$

$$\begin{gathered} y_i=a{x}_i+b \\ \Rightarrow \frac{\sum _{}{y}_i}{n}=\frac{\sum _{}\left(a{x}_i+b\right)}{n} \\ \Rightarrow \frac{\sum _{}{y}_i}{n}=a\frac{\sum _{}{x}_i}{n}+\frac{\sum _{}b}{n} \\ \Rightarrow \overline{y_i}=a\overline{x_i}+b \\ \Rightarrow 55=48a+b.....\left(1\right) \end{gathered}$$

Now,

Standard deviation of y_i = Standard deviation of $a{x}_i+b$

$$\begin{gathered} \Rightarrow {\sigma }_{y_i}=a\times {\sigma }_{x_i} \\ \Rightarrow 15=12a \\ \Rightarrow a=\frac{15}{12}=1.25 \end{gathered}$$

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.

Hence, the correct answer is option (a).