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  • There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test: Marks 0 Frequency x-2

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 } & \mathrm{x}-2 & \mathrm{x} & \mathrm{x}^{2} & (\mathrm{x}+1)^{2} & 2 \mathrm{x} & \mathrm{x}+1 \\ \hline \end{array}

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

 

 

Answers (1)

It is given that there are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given.

We have to find the mean and standard deviation of the marks.

It is given that there are 60 students in the class, so

\\ \Sigma f_{i}=60~ \\\\ \\ \left( x-2 \right) +x+x^{2}+ \left( x+1 \right) ^{2}+2x+x+1=60~ \\\\ \\ 5x-1+x^{2}+x^{2}+2x+1=60 \\\\ \\ 2x^{2}+7x=60 \\\\ \\ \text{On factorising 2}x^{2}+7x-60=0 \\\\ \\ \text{we get } \left( 2x+15 \right) \left( x-4 \right) =0 \\\\ \\ 2x=-15 or x=4 \\\\ \\ \text{Given~that x is a positive number, so x can take 4 as the only value .} \\\\ \\ \text{And let assumed mean}, a=3 \\\\ \\ \text{Now put x=4 and a=3} \text{ in the frequency distribution table and other columns after} \\\\ \\ \text{calculations, we get } \\\\

\begin{array}{|l|l|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & d_{i}=x_{i}-a_{i} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 0 & \mathrm{x}-2=4-2=2 & -3 & -6 & 18 \\ \hline 1 & \mathrm{X}=4 & -2 & -8 & 16 \\ \hline 2 & x^{2}=16 & -1 & -16 & 16 \\ \hline 3 & (x+1)^{2} & 0 & 0 & 0 \\ & =(4+1)^{2}=25 & & & \\ \hline 4 & 2 \mathrm{x}=2 * 4=8 & 1 & 4 & 8 \\ \hline 5 & \mathrm{X}+1=4+1=5 & 2 & 10 & 20 \\ \hline \text { Total } & \mathrm{N}=60 & 277 & =-12 & 78 \\ \hline \end{array}

And we know that standard deviation is  \\\\ \sigma = \sqrt {\frac{ \Sigma f_{i}d_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}d_{i}}{n} \right) ^{2}}~ \\\\


\\ \text{Substituting the values from the above table } \sigma =\sqrt {\frac{78}{60}- \left( \frac{-12}{60} \right) ^{2}} \\\\ \\ =\sqrt {1.3- \left( 0.2 \right) ^{2}}=\sqrt {1.3-0.04}=1.12~ \\\\ \\ \text{Hence, the standard deviation is 1.12 } \\\\ \\ \text{ Now mean is }\overline{x}=A+\frac{ \Sigma f_{i}d_{i}}{N}~ =3+ \left( -\frac{12}{60} \right) =3-\frac{1}{5}=\frac{14}{5}=2.8~~ \\\\ \\ \text{Hence, the mean and standard deviation of the marks are 2.8 and 1.12 respectively} \\\\

 

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