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Home » The mean and standard deviation of six observations are and , respectively. If each observation is multiplied by , find the new mean and new standard deviation of the resulting observations.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
CBSE | Class 11 | Maths | Miscellaneous Exercise | NCERT
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

We are given that,

The mean of six observations = 8.

The standard deviation of six observations = 4.

Let us consider the six observations as {x_1}, {x_2}, {x_3},  {x_4}, {x_5} and {x_6}.

The mean of six observations is given by,

\bar x = \frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8

When each observation is multiplied by 3 then,

    \[{y_i} = 3{x_i}\]

So,

    \[{x_i} = \frac{1}{3}{y_i}\]

, for i = 1 to 6      …… (1)

Now, the new mean becomes,

\bar y = \frac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}

\bar y = \frac{{3\left( {{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}} \right)}}{6}

\bar y = 3 \times 8

\bar y = 24

Also, we have Standard deviation,

\sigma = \sqrt {\frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} - \overline {\text{x}} } \right)}^2}} }

Substituting the values and squaring on both sides,

{\sigma ^2} = \frac{1}{6}\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} - \overline {\text{x}} } \right)}^2}}

{\left( 4 \right)^2} = \frac{1}{6}\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} - \overline {\text{x}} } \right)}^2}}

\sum\limits_{{\text{i}} = 1}^6 {{{\left( {{{\text{x}}_{\text{i}}} - \overline {\text{x}} } \right)}^2}} = 96                       …… (2)

From equations (1) and (2),

\sum\limits_{{\text{i}} = 1}^6 {{{\left( {\frac{1}{3}{y_{\text{i}}} - \frac{1}{3}\overline y } \right)}^2}} = 96

\sum\limits_{{\text{i}} = 1}^6 {\left( {\frac{1}{3}{y_{\text{i}}} - \frac{1}{3}\overline y } \right)} = 864

So, the new variance of the resulting observation = \frac{1}{6} \times 864

= 144

Hence, the new standard deviation of the resulting observation = \sqrt {144}

= 12

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