Question

Q1. To draw a histogram to represent the following frequency distribution:

Class Interval	5 – 10	10 – 15	15 – 25	25 – 45	45 – 75
Frequency	6	12	10	8	15

the adjusted frequency for the class 25 ​– 45 is :
(A) 6 (B) 5 (C) 3 (D) 2

Q2. If the mean of the observations:
x, x + 3, x + 5, x + 7, x + 10
is 9, the mean of the last three observations is
(A) $10\frac{1}{3}$ (B) $10\frac{2}{3}$ (C) $11\frac{1}{3}$ (D) $11\frac{2}{3}$

Answer 1

Dear student
$$\begin{gathered} \mathrm{Mean}\mathrm{of}\mathrm{observations}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{observations}}{\mathrm{Number}\mathrm{of}\mathrm{observations}} \\ \mathrm{Now},\mathrm{the}\mathrm{given}\mathrm{observations}\mathrm{are}: \\ \mathrm{x},\mathrm{x}+3,\mathrm{x}+5,\mathrm{x}+7,\mathrm{x}+10 \\ \mathrm{Now},\mathrm{mean}=\frac{\mathrm{x}+\mathrm{x}+3+\mathrm{x}+5+\mathrm{x}+7+\mathrm{x}+10}{5} \\ \Rightarrow 9=\frac{5\mathrm{x}+25}{5} \\ \Rightarrow 5\mathrm{x}+25=45 \\ \Rightarrow 5\mathrm{x}=20 \\ \Rightarrow \mathrm{x}=4 \\ \mathrm{So},\mathrm{the}\mathrm{observations}\mathrm{are}: \\ 4,7,9,11,14 \\ \mathrm{Now},\mathrm{the}\mathrm{maen}\mathrm{of}\mathrm{last}\mathrm{threre}\mathrm{observations}\mathrm{is} \\ \mathrm{Mean}=\frac{9+11+14}{3} \\ \Rightarrow \mathrm{x}=\frac{34}{3} \\ \Rightarrow \mathrm{x}=11\frac{1}{3} \end{gathered}$$
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