There are 60 students in a class. following is frequency distribution of marks obtained by students

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13.Statistics

hard

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Marks } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & x-2 & x & x^{2} & (x+1)^{2} & 2 x & x+1 \\ \hline \end{array}$

where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

Option A

Option B

Option C

Option D

Solution

Sum of frequencies,

$x-2+x+x^{2}+(x+1)^{2}+2 x+x+1=60$

$2 x^{2}+7 x-60=0$

$(2 x+15)(x-4)=0$

$x=4$

$\begin{array}{|c|c|c|c|c|} \hline x _{ i } & f_{i} & d_{i}=x_{i}-3 & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 0 & 2 & -3 & -6 & 18 \\ \hline 1 & 4 & -2 & -8 & 16 \\ \hline 2 & 16 & -1 & -16 & 16 \\ \hline A=3 & 25 & 0 & 0 & 0 \\ \hline 4 & 8 & 1 & 8 & 8 \\ \hline 5 & 5 & 2 & 10 & 20 \\ \hline \text { Total } & \Sigma f_{i}=60 & & \Sigma f_{i}=-12 & \Sigma f_{i} d_{i}^{2}=78 \\ \hline \end{array}$

Mean $=A+\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}=3+\left(\frac{-12}{60}\right)=2.8$

Standard Deviation,

$\sigma$=$\sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\right)^{2}}=\sqrt{\frac{78}{60}-\left(\frac{-12}{60}\right)^{2}}=\sqrt{1.3-0.04}=\sqrt{1.26}=1.12$

Standard 11

Mathematics

The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are…

medium

If each of the observation $x_{1}, x_{2}, \ldots ., x_{n}$ is increased by $'a'$ where $a$ is a negative or positive number, show that the variance remains unchanged.

hard

The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted a and $b$ are respectively mean and variance of remaining $6$ observation, then $a+3 b-5$ is equal to $……….$.

hard

(JEE MAIN-2023)

Let the mean of the data

$X$ $1$ $3$ $5$ $7$ $9$

$(f)$ $4$ $24$ $28$ $\alpha$ $8$

be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $……….$.

hard

(JEE MAIN-2023)

Find the mean and variance for the following frequency distribution.

Classes $0-10$ $10-20$ $20-30$ $30-40$ $40-50$

Frequencies $5$ $8$ $15$ $16$ $6$

hard

Classes	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequencies	$5$	$8$	$15$	$16$	$6$

Solution

Similar Questions

The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are…

If each of the observation $x_{1}, x_{2}, \ldots ., x_{n}$ is increased by $'a'$ where $a$ is a negative or positive number, show that the variance remains unchanged.

The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted a and $b$ are respectively mean and variance of remaining $6$ observation, then $a+3 b-5$ is equal to $……….$.

Let the mean of the data $X$ $1$ $3$ $5$ $7$ $9$ $(f)$ $4$ $24$ $28$ $\alpha$ $8$ be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $……….$.

Find the mean and variance for the following frequency distribution. Classes $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ Frequencies $5$ $8$ $15$ $16$ $6$

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Let the mean of the data

$X$ $1$ $3$ $5$ $7$ $9$

$(f)$ $4$ $24$ $28$ $\alpha$ $8$

be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $……….$.

Find the mean and variance for the following frequency distribution.

Classes $0-10$ $10-20$ $20-30$ $30-40$ $40-50$

Frequencies $5$ $8$ $15$ $16$ $6$