The marks scored by the students in slip test are given below:
$x$ 4 6 8 10 12
$f$ 7 3 5 9 5

Find the standard deviation of these marks.

Question

The marks scored by the students in slip test are given below:$x$4681012$f$73595Find the standard deviation of these marks.

Vedantu Content Team · Accepted Answer

Hint: To obtain the standard deviation of data we will use the standard deviation formula for grouped data. Firstly we will find the mean of the data then we will subtract the mean with the marks in each column and add them. Finally we will find the standard deviation by dividing the obtained difference in mean and actual marks by total students and get our desired answer.Complete step by step answer:The table given us is as below:$x$$f$476385109125 Now we have to find the mean marks of the data so we will use the below formula:Mean Marks$=\overline{x}=\dfrac{\sum{fx}}{\sum{f}}$….$\left( 1 \right)$So we will get the values from our table as follows:$x$$f$ $fx$47$4\times 7=28$63$6\times 3=18$85$8\times 5=40$109$10\times 9=90$125$12\times 5=60$$\sum{f}=29$$\sum{fx}=236$  Put value from above table in equation (1) we get,$\begin{align}  & \overline{x}=\dfrac{236}{29} \\  & \Rightarrow \overline{x}=8.14 \\ \end{align}$So$\overline{x}=8.14\approx 8$….$\left( 2 \right)$Now we will use the below formula to find the standard deviation:Standard deviation $=\sigma =\sqrt{\dfrac{\sum{f{{\left( \overline{x}-x \right)}^{2}}}}{\sum{f}}}$ …..$\left( 3 \right)$Now we will use the table to find the value for the above formula as below:$x$$f$ $fx$$\overline{x}-x$${{\left( \overline{x}-x \right)}^{2}}$$f{{\left( \overline{x}-x \right)}^{2}}$47$4\times 7=28$$4-8=-4$${{\left( -4 \right)}^{2}}=16$$7\times 16=112$63$6\times 3=18$$6-8=-2$${{\left( -2 \right)}^{2}}=4$$3\times 4=12$85$8\times 5=40$$8-8=0$$0$$5\times 0=0$109$10\times 9=90$$10-8=2$${{\left( 2 \right)}^{2}}=4$$9\times 4=36$125$12\times 5=60$$12-8=4$${{\left( 4 \right)}^{2}}=16$$5\times 16=80$$\sum{f}=29$$\sum{fx}=236$$\sum{f{{\left( \overline{x}-x \right)}^{2}}=240}$Put the value from above table in equation (3) we get,$\begin{align}  & \sigma =\sqrt{\dfrac{240}{29}} \\  & \sigma =2.877 \\ \end{align}$$\sigma =2.88$Hence standard deviation of given marks is $2.88$Note:  Grouped data is formed by aggregating individual observations of a variable into groups so that a frequency distribution of these groups can be used to summarize it. Standard deviation is used to calculate the amount of variation or dispersion of a set of values. If the value of standard deviation is low that means the values are close to the mean and if the standard deviation is high that means the values are spread out over a wider range.

$x$	$f$	$fx$
4	7	$4\times 7=28$
6	3	$6\times 3=18$
8	5	$8\times 5=40$
10	9	$10\times 9=90$
12	5	$12\times 5=60$
	$\sum{f}=29$	$\sum{fx}=236$

$x$	$f$	$fx$	$\overline{x}-x$	${{\left( \overline{x}-x \right)}^{2}}$	$f{{\left( \overline{x}-x \right)}^{2}}$
4	7	$4\times 7=28$	$4-8=-4$	${{\left( -4 \right)}^{2}}=16$	$7\times 16=112$
6	3	$6\times 3=18$	$6-8=-2$	${{\left( -2 \right)}^{2}}=4$	$3\times 4=12$
8	5	$8\times 5=40$	$8-8=0$	$0$	$5\times 0=0$
10	9	$10\times 9=90$	$10-8=2$	${{\left( 2 \right)}^{2}}=4$	$9\times 4=36$
12	5	$12\times 5=60$	$12-8=4$	${{\left( 4 \right)}^{2}}=16$	$5\times 16=80$
	$\sum{f}=29$	$\sum{fx}=236$			$\sum{f{{\left( \overline{x}-x \right)}^{2}}=240}$