Mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking

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13.Statistics

hard

The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12$ . If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to$...................$

$2521$

$3562$

$1245$

$2356$

(JEE MAIN-2024)

Solution

Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$

We have

$\mu^{\prime}=\frac{\Sigma x_i}{15}=12 \Rightarrow \Sigma x_i=180$

As per given information correct $\Sigma x_i=180-10+12$

$\Rightarrow \mu(\text { correct mean })=\frac{182}{15}$

Also

$ \sigma^{\prime}=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295 $

$\text { Correct } \Sigma \mathrm{x}_{\mathrm{i}}^2=2295-100+144=2339 $

$ \sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}$

Required value

$ =15\left(\mu+\mu^2+\sigma^2\right) $

$ =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) $

$ =15\left(\frac{182}{15}+\frac{2339}{15}\right) $

$ =2521$

Standard 11

Mathematics

In an experiment with $15$ observations on $x$, the following results were available $\sum {x^2} = 2830$, $\sum x = 170$. On observation that was $20$ was found to be wrong and was replaced by the correct value $30$. Then the corrected variance is..

hard

(AIEEE-2003)

The mean and variance of the marks obtained by the students in a test are $10$ and $4$ respectively. Later, the marks of one of the students is increased from $8$ to $12$ . If the new mean of the marks is $10.2.$ then their new variance is equal to :

hard

(JEE MAIN-2023)

Suppose values taken by a variable $x$ are such that $a \le {x_i} \le b$, where ${x_i}$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, …n.$ Then..

normal

From the data given below state which group is more variable, $A$ or $B$ ?

Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Group $A$ $9$ $17$ $32$ $33$ $40$ $10$ $9$

Group $B$ $10$ $20$ $30$ $25$ $43$ $15$ $7$

hard

Let $x_1, x_2, x_3, x_4, ………. , x_n$ be $n$ observations and let $\bar x$ be their arithmetic mean and $\sigma ^2$ be their variance.

Statement $-1$ : Variance of observations $2x_1, 2x_2, 2x_3, ……, 2x_n$ is $4\sigma ^2$ .

Statement $-2$ : Arithmetic mean of $2x _1, 2x_2, 2x_3, ……, 2x_n$ is $4\bar x$ .

normal

Marks	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Group $A$	$9$	$17$	$32$	$33$	$40$	$10$	$9$
Group $B$	$10$	$20$	$30$	$25$	$43$	$15$	$7$

Solution

Similar Questions

In an experiment with $15$ observations on $x$, the following results were available $\sum {x^2} = 2830$, $\sum x = 170$. On observation that was $20$ was found to be wrong and was replaced by the correct value $30$. Then the corrected variance is..

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From the data given below state which group is more variable, $A$ or $B$ ?

Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Group $A$ $9$ $17$ $32$ $33$ $40$ $10$ $9$

Group $B$ $10$ $20$ $30$ $25$ $43$ $15$ $7$