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Home » The mean and standard deviation of observations are found to be and , respectively. On rechecking, it was found that an observation was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by .
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12.
CBSE | Class 11 | Maths | Miscellaneous Exercise | NCERT
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12.

(i) When the wrong item is omitted,

We are given,

The number of observations, n = 20.

The mean before rechecking = 10.

The standard deviation before rechecking = 2.

Mean is given by,

\overline {\text{X}} = \frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{20} {{{\text{X}}_1}}

10 = \frac{1}{{20}}\sum\limits_{{\text{i}} = 1}^{20} {{{\text{X}}_1}}

\sum\limits_{{\text{i}} = 1}^{20} {{{\text{X}}_1}} = 200

So, the sum of observations before rechecking = 200

Thus, the correct sum of observations = 200 - 8

= 192

Hence, the correct mean = \frac{{correct{\text{ }}sum}}{{19}}

= \frac{{192}}{{19}}

= 10.1

Also, the standard deviation is given by,

\sigma = \sqrt {\frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{\text{n}} {{{\text{X}}_1}} - \frac{1}{{{{\text{n}}^2}}}{{\left( {\sum\limits_{i = 1}^{\text{n}} {\text{X}} } \right)}^2}}

    \[2 = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {X_1^2} - {{(\bar X)}^2}} \]

Squaring on both sides and simplifying we get,

    \[4 = \frac{1}{{20}}{\text{ Incorrect }}\sum\limits_{i = 1}^n {X_1^2} - 100{\text{ }}\]

    \[{\text{Incorrect }}\sum\limits_{i = 1}^n {X_1^2} = 2080\]

Now,

    \[Correct\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2} = Incorrect\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2} - {(8)^2}\]

    \[ = 2080 - 64\]

    \[ = 2016\]

Therefore, the correct standard deviation is,

= \sqrt {\frac{{{\text{ Correct }}\sum {{\text{X}}_1^2} }}{{\text{n}}} - {{({\text{ Correct Mean }})}^2}}

= \sqrt {\frac{{2016}}{{19}} - {{(10.1)}^2}}

= \sqrt {1061.1 - 102.1}

= 2.02

(ii) when the wrong item is replaced by 12,

We are given,

The sum of observations before rechecking = 200

The correct sum of observations = 200 - 8 + 12

= 204

So, the correct mean = \frac{{correct{\text{ }}sum}}{{20}}

= \frac{{204}}{{20}}

= 10.2

Also, the standard deviation is given by,

\sigma = \sqrt {\frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{\text{n}} {{{\text{X}}_1}} - \frac{1}{{{{\text{n}}^2}}}{{\left( {\sum\limits_{i = 1}^{\text{n}} {\text{X}} } \right)}^2}}

    \[2 = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {X_1^2} - {{(\bar X)}^2}} \]

Squaring on both sides and simplifying we get,

    \[4 = \frac{1}{{20}}{\text{ Incorrect }}\sum\limits_{i = 1}^n {X_1^2} - 100{\text{ }}\]

    \[{\text{Incorrect }}\sum\limits_{i = 1}^n {X_1^2} = 2080\]

Now,

    \[Correct\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2} = Incorrect\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2} - {(8)^2} + {(12)^2}\]

    \[ = 2080 - 64 + 144\]

    \[ = 2160\]

Therefore, the correct standard deviation is,

= \sqrt {\frac{{{\text{ Correct }}\sum {{\text{X}}_1^2} }}{{\text{n}}} - {{({\text{ Correct Mean }})}^2}}

= \sqrt {\frac{{2160}}{{20}} - {{(10.2)}^2}}

= \sqrt {108 - 104.04}

= 1.98

i

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