Class 10: Statistics – Very Short Answer Type Questions

$\displaystyle \textbf{Question 1: }\text{Define mean.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Mean is the sum of all observations divided by the total number of observations.}$
$\displaystyle \bar{x}=\frac{\sum x_i}{n}$
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$\displaystyle \textbf{Question 2: }\text{What is the algebraic sum of deviations of a frequency}$
$\displaystyle \text{distribution about its mean?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The algebraic sum of deviations about the mean is zero.}$
$\displaystyle \sum f_i(x_i-\bar{x})=0$
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$\displaystyle \textbf{Question 3: }\text{Which measure of central tendency is given by the }x\text{-coordinate}$
$\displaystyle \text{of the point of intersection of the more than ogive and less than ogive?}$
$\displaystyle \hfill\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Median.}$
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$\displaystyle \textbf{Question 4: }\text{Write the empirical relation between mean, mode and median.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Mode}=3\text{ Median}-2\text{ Mean}.$
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$\displaystyle \textbf{Question 5: }\text{Which measure of central tendency can be determined graphically?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Median and mode can be determined graphically.}$
$\displaystyle \text{Median is obtained using ogives and mode is obtained using a histogram.}$
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$\displaystyle \textbf{Question 6: }\text{Find the mode of the following frequency distribution:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Class} & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 \\ \hline \text{Frequency} & 8 & 7 & 12 & 5 & 3 \\ \hline \end{array}$
$\displaystyle \hfill\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The maximum frequency is }12,\text{ so }40-60\text{ is the modal class.}$
$\displaystyle l=40,\quad h=20,\quad f=12,\quad f_1=7,\quad f_2=5$
$\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h$
$\displaystyle =40+\frac{12-7}{2(12)-7-5}\times20$
$\displaystyle =40+\frac{5}{12}\times20=48.33\text{ (approximately)}$
$\displaystyle \therefore\ \text{Mode}=48.33\text{ (approximately).}$
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$\displaystyle \textbf{Question 7: }\text{If mode of the following frequency distribution is }55,\text{ then}$
$\displaystyle \text{find the value of }x.$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Class} & 0-15 & 15-30 & 30-45 & 45-60 & 60-75 & 75-90 \\ \hline \text{Frequency} & 10 & 7 & x & 15 & 10 & 12 \\ \hline \end{array}$
$\displaystyle \hfill\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since mode }=55,\text{ the modal class is }45-60.$
$\displaystyle l=45,\quad h=15,\quad f=15,\quad f_1=x,\quad f_2=10$
$\displaystyle \text{Mode}=l+\frac{f-f_1}{2f-f_1-f_2}\times h$
$\displaystyle 55=45+\frac{15-x}{30-x-10}\times15$
$\displaystyle 10=\frac{15(15-x)}{20-x}$
$\displaystyle 2(20-x)=3(15-x)$
$\displaystyle 40-2x=45-3x$
$\displaystyle \therefore\ x=5.$
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$\displaystyle \textbf{Question 8: }\text{Write the modal class for the following frequency distribution:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Class interval} & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 \\ \hline \text{Frequency} & 30 & 35 & 75 & 40 & 30 & 15 \\ \hline \end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The maximum frequency is }75,\text{ corresponding to the class }20-25.$
$\displaystyle \therefore\ \text{Modal class}=20-25.$
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$\displaystyle \textbf{Question 9: }\text{Write the median class for the following frequency distribution:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Class interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{Frequency} & 5 & 8 & 7 & 12 & 28 & 20 & 10 & 10 \\ \hline \end{array}$
$\displaystyle \text{Answer:}$
$\displaystyle N=5+8+7+12+28+20+10+10=100$
$\displaystyle \frac{N}{2}=\frac{100}{2}=50$
$\displaystyle \begin{array}{|c|c|} \hline \text{Class interval} & \text{Cumulative frequency} \\ \hline 0-10 & 5 \\ \hline 10-20 & 13 \\ \hline 20-30 & 20 \\ \hline 30-40 & 32 \\ \hline 40-50 & 60 \\ \hline 50-60 & 80 \\ \hline 60-70 & 90 \\ \hline 70-80 & 100 \\ \hline \end{array}$
$\displaystyle \text{The cumulative frequency just greater than }50\text{ is }60.$
$\displaystyle \therefore\ \text{Median class}=40-50.$
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$\displaystyle \textbf{Question 10: }\text{In the graphical representation of a frequency distribution,}$
$\displaystyle \text{if the distance between mode and mean is }k\text{ times the distance between}$
$\displaystyle \text{median and mean, then write the value of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Mode}=3\text{ Median}-2\text{ Mean}$
$\displaystyle \Rightarrow\text{Mode}-\text{Mean}=3(\text{Median}-\text{Mean})$
$\displaystyle \therefore\ k=3.$
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$\displaystyle \textbf{Question 11: }\text{Find the class marks of classes }10-25\text{ and }35-55.$
$\displaystyle \hfill\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Class mark}=\frac{\text{Lower limit}+\text{Upper limit}}{2}$
$\displaystyle \text{For }10-25,\quad\text{class mark}=\frac{10+25}{2}=17.5$
$\displaystyle \text{For }35-55,\quad\text{class mark}=\frac{35+55}{2}=45$
$\displaystyle \therefore\ \text{The class marks are }17.5\text{ and }45.$
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$\displaystyle \textbf{Question 12: }\text{Write the median class of the following distribution:}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Classes} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 \\ \hline \text{Frequency} & 4 & 4 & 8 & 10 & 12 & 8 & 4 \\ \hline \end{array}$
$\displaystyle \hfill\text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle N=4+4+8+10+12+8+4=50$
$\displaystyle \frac{N}{2}=\frac{50}{2}=25$
$\displaystyle \begin{array}{|c|c|} \hline \text{Class interval} & \text{Cumulative frequency} \\ \hline 0-10 & 4 \\ \hline 10-20 & 8 \\ \hline 20-30 & 16 \\ \hline 30-40 & 26 \\ \hline 40-50 & 38 \\ \hline 50-60 & 46 \\ \hline 60-70 & 50 \\ \hline \end{array}$
$\displaystyle \text{The cumulative frequency just greater than }25\text{ is }26.$
$\displaystyle \therefore\ \text{Median class}=30-40.$
$\displaystyle \\$