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.
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.
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,
$\sigma$=$\sqrt{\frac{\Sigma f_{i} d_{i}^{2}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\right)^{2}}=\sqrt{\frac{78}{60}-\left(\frac{-12}{60}\right)^{2}}=\sqrt{1.3-0.04}=\sqrt{1.26}=1.12$
Similar Questions
If the mean of the frequency distribution
Class:
$0-10$
$10-20$
$20-30$
$30-40$
$40-50$
Frequency
$2$
$3$
$x$
$5$
$4$
is $28$ , then its variance is $........$.
If the mean of the frequency distribution
| Class: | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
| Frequency | $2$ | $3$ | $x$ | $5$ | $4$ |
is $28$ , then its variance is $........$.
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$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.
The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
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Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
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If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to
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