\displaystyle \textbf{Question 1: }\text{If the heights of }5\text{ persons are }140\text{ cm},150\text{ cm},152\text{ cm},
\displaystyle 158\text{ cm and }161\text{ cm respectively, find the mean height.}
\displaystyle \text{Answer:}
\displaystyle n=5
\displaystyle \text{Sum of the heights}=140+150+152+158+161=761\text{ cm}
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{761}{5}=152.2\text{ cm}
\displaystyle \therefore \text{The mean height is }152.2\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the mean of }994,996,998,1002\text{ and }1000.
\displaystyle \text{Answer:}
\displaystyle n=5
\displaystyle \text{Sum of the numbers}=994+996+998+1002+1000=4990
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{4990}{5}=998
\displaystyle \therefore \text{The mean is }998.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the mean of the first five natural numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{The first five natural numbers are }1,2,3,4,5.
\displaystyle n=5
\displaystyle \text{Sum of the numbers}=1+2+3+4+5=15
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{15}{5}=3
\displaystyle \therefore \text{The mean of the first five natural numbers is }3.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the mean of all the factors of }10.
\displaystyle \text{Answer:}
\displaystyle \text{The factors of }10\text{ are }1,2,5,10.
\displaystyle n=4
\displaystyle \text{Sum of the factors}=1+2+5+10=18
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{18}{4}=4.5
\displaystyle \therefore \text{The mean of all the factors of }10\text{ is }4.5.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the mean of the first }10\text{ even natural numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{The first }10\text{ even natural numbers are }2,4,6,8,10,12,14,16,18,20.
\displaystyle n=10
\displaystyle \text{Sum of the numbers}=2+4+6+8+10+12+14+16+18+20=110
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{110}{10}=11
\displaystyle \therefore \text{The mean of the first }10\text{ even natural numbers is }11.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the mean of }x,x+2,x+4,x+6\text{ and }x+8.
\displaystyle \text{Answer:}
\displaystyle n=5
\displaystyle \text{Sum of the observations}=x+(x+2)+(x+4)+(x+6)+(x+8)
\displaystyle =5x+20
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{5x+20}{5}=x+4
\displaystyle \therefore \text{The mean is }x+4.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the mean of the first five multiples of }3.
\displaystyle \text{Answer:}
\displaystyle \text{The first five multiples of }3\text{ are }3,6,9,12,15.
\displaystyle n=5
\displaystyle \text{Sum of the numbers}=3+6+9+12+15=45
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{45}{5}=9
\displaystyle \therefore \text{The mean of the first five multiples of }3\text{ is }9.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The weights (in kg) of }10\text{ new born babies in a hospital on a particular}
\displaystyle \text{day are }3.4,3.6,4.2,4.5,3.9,4.1,3.8,4.5,4.4,3.6.\text{ Find the mean.}
\displaystyle \text{Answer:}
\displaystyle n=10
\displaystyle \text{Sum of the weights}=3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6=40\text{ kg}
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{40}{10}=4\text{ kg}
\displaystyle \therefore \text{The mean weight of the babies is }4\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The percentage of marks obtained by students of a class in Mathematics are}
\displaystyle 64,36,47,23,0,19,81,93,72,35,3,1.\text{ Find their mean.}
\displaystyle \text{Answer:}
\displaystyle n=12
\displaystyle \text{Sum of the percentages}=64+36+47+23+0+19+81+93+72+35+3+1=474
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{474}{12}=39.5
\displaystyle \therefore \text{The mean percentage is }39.5.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The numbers of children in the following families are }2,4,3,4,2,0,3,5,1,1,5.
\displaystyle \text{Find the mean number of children per family.}
\displaystyle \text{Answer:}
\displaystyle n=11
\displaystyle \text{Sum of the numbers}=2+4+3+4+2+0+3+5+1+1+5=30
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{30}{11}\approx2.73
\displaystyle \therefore \text{The mean number of children per family is }2.73\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }M\text{ is the mean of }x_1,x_2,x_3,x_4,x_5\text{ and }x_6,\text{ prove that}
\displaystyle (x_1-M)+(x_2-M)+(x_3-M)+(x_4-M)+(x_5-M)+(x_6-M)=0.
\displaystyle \text{Answer:}
\displaystyle M=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}
\displaystyle \Rightarrow 6M=x_1+x_2+x_3+x_4+x_5+x_6
\displaystyle \Rightarrow x_1+x_2+x_3+x_4+x_5+x_6-6M=0
\displaystyle \Rightarrow (x_1-M)+(x_2-M)+(x_3-M)+(x_4-M)+(x_5-M)+(x_6-M)=0
\displaystyle \therefore \sum_{i=1}^{6}(x_i-M)=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The duration of sunshine (in hours) in Amritsar during the first }10\text{ days of}
\displaystyle \text{August }1997\text{ is }9.6,5.2,3.5,1.5,1.6,2.4,2.6,8.4,10.3,10.9.
\displaystyle \text{(i) Find the mean. (ii) Verify that }\sum_{i=1}^{10}(x_i-\overline{x})=0.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }n=10
\displaystyle \text{Sum of the sunshine durations}=9.6+5.2+3.5+1.5+1.6+2.4+2.6+8.4+10.3+10.9=56
\displaystyle \text{Mean }(\overline{x})=\frac{\sum x_i}{n}
\displaystyle =\frac{56}{10}=5.6\text{ hours}
\displaystyle \text{(ii) }\sum_{i=1}^{10}(x_i-\overline{x})
\displaystyle =(9.6-5.6)+(5.2-5.6)+(3.5-5.6)+(1.5-5.6)+(1.6-5.6)+(2.4-5.6)
\displaystyle +(2.6-5.6)+(8.4-5.6)+(10.3-5.6)+(10.9-5.6)
\displaystyle =4-0.4-2.1-4.1-4-3.2-3+2.8+4.7+5.3
\displaystyle =0
\displaystyle \therefore \sum_{i=1}^{10}(x_i-\overline{x})=0.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Explain, by taking a suitable example, how the arithmetic mean alters by}
\displaystyle \text{(i) adding a constant }k\text{ to each term, (ii) subtracting a constant }k\text{ from each term,}
\displaystyle \text{(iii) multiplying each term by a constant }k\text{ and (iv) dividing each term by a}
\displaystyle \text{non-zero constant }k.
\displaystyle \text{Answer:}
\displaystyle \text{Let the mean of }x_1,x_2,\ldots,x_n\text{ be }\overline{X}.
\displaystyle \therefore \overline{X}=\frac{1}{n}\sum_{i=1}^{n}x_i

\displaystyle \text{(i) Let }\overline{X_1}\text{ be the mean of }x_1+k,x_2+k,\ldots,x_n+k.
\displaystyle \overline{X_1}=\frac{(x_1+k)+(x_2+k)+\cdots+(x_n+k)}{n}
\displaystyle =\frac{x_1+x_2+\cdots+x_n+nk}{n}
\displaystyle =\frac{1}{n}\sum_{i=1}^{n}x_i+k=\overline{X}+k
\displaystyle \therefore \text{Adding }k\text{ to each observation increases the mean by }k.

\displaystyle \text{(ii) Let }\overline{X_1}\text{ be the mean of }x_1-k,x_2-k,\ldots,x_n-k.
\displaystyle \overline{X_1}=\frac{(x_1-k)+(x_2-k)+\cdots+(x_n-k)}{n}
\displaystyle =\frac{x_1+x_2+\cdots+x_n-nk}{n}
\displaystyle =\frac{1}{n}\sum_{i=1}^{n}x_i-k=\overline{X}-k
\displaystyle \therefore \text{Subtracting }k\text{ from each observation decreases the mean by }k.

\displaystyle \text{(iii) Let }\overline{X_1}\text{ be the mean of }kx_1,kx_2,\ldots,kx_n.
\displaystyle \overline{X_1}=\frac{kx_1+kx_2+\cdots+kx_n}{n}
\displaystyle =\frac{k}{n}(x_1+x_2+\cdots+x_n)
\displaystyle =k\left(\frac{1}{n}\sum_{i=1}^{n}x_i\right)=k\overline{X}
\displaystyle \therefore \text{Multiplying each observation by }k\text{ multiplies the mean by }k.

\displaystyle \text{(iv) Let }\overline{X_1}\text{ be the mean of }\frac{x_1}{k},\frac{x_2}{k},\ldots,\frac{x_n}{k},\ k\ne0.
\displaystyle \overline{X_1}=\frac{\frac{x_1}{k}+\frac{x_2}{k}+\cdots+\frac{x_n}{k}}{n}
\displaystyle =\frac{1}{kn}(x_1+x_2+\cdots+x_n)
\displaystyle =\frac{1}{k}\left(\frac{1}{n}\sum_{i=1}^{n}x_i\right)=\frac{\overline{X}}{k}
\displaystyle \therefore \text{Dividing each observation by }k\text{ divides the mean by }k.

\displaystyle \text{For example, consider }2,4,6.\text{ Their mean is }\frac{2+4+6}{3}=4.
\displaystyle \text{Taking }k=2,\text{ the means after adding, subtracting, multiplying and dividing by }2
\displaystyle \text{are respectively }6,2,8\text{ and }2,\text{ i.e. }4+2,\ 4-2,\ 4\times2\text{ and }\frac{4}{2}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The mean of marks scored by }100\text{ students was found to be }40.
\displaystyle \text{Later it was discovered that a score of }53\text{ was misread as }83.\text{ Find the correct mean.}
\displaystyle \text{Answer:}
\displaystyle n=100,\qquad \text{Incorrect mean}=40
\displaystyle \text{Incorrect total marks}=40\times100=4000
\displaystyle \text{The score }83\text{ should be }53.
\displaystyle \text{Correct total marks}=4000-83+53=3970
\displaystyle \text{Correct mean}=\frac{3970}{100}=39.7
\displaystyle \therefore \text{The correct mean is }39.7.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The traffic police recorded the speeds (in km/hr) of }10\text{ motorists as}
\displaystyle 47,53,49,60,39,42,55,57,52,48.\text{ Later it was found that the recording}
\displaystyle \text{instrument recorded }5\text{ km/hr less in each case. Find the correct average speed.}
\displaystyle \text{Answer:}
\displaystyle n=10
\displaystyle \text{Sum of the recorded speeds}=47+53+49+60+39+42+55+57+52+48=502\text{ km/hr}
\displaystyle \text{Recorded mean}=\frac{502}{10}=50.2\text{ km/hr}
\displaystyle \text{Since each recorded speed is }5\text{ km/hr less than the actual speed,}
\displaystyle \text{Correct mean}=50.2+5=55.2\text{ km/hr}
\displaystyle \therefore \text{The correct average speed of the motorists is }55.2\text{ km/hr}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The mean of five numbers is }27.\text{ If one number is excluded, their mean is}
\displaystyle 25.\text{ Find the excluded number.}
\displaystyle \text{Answer:}
\displaystyle \text{Total of the five numbers}=5\times27=135
\displaystyle \text{Total of the remaining four numbers}=4\times25=100
\displaystyle \text{Excluded number}=135-100=35
\displaystyle \therefore \text{The excluded number is }35.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The mean weight per student in a group of }7\text{ students is }55\text{ kg. The}
\displaystyle \text{individual weights of }6\text{ of them are }52,54,55,53,56\text{ and }54\text{ kg. Find the}
\displaystyle \text{weight of the seventh student.}
\displaystyle \text{Answer:}
\displaystyle \text{Total weight of }7\text{ students}=7\times55=385\text{ kg}
\displaystyle \text{Total weight of the given }6\text{ students}=52+54+55+53+56+54=324\text{ kg}
\displaystyle \text{Weight of the seventh student}=385-324=61\text{ kg}
\displaystyle \therefore \text{The weight of the seventh student is }61\text{ kg.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The mean of }8\text{ numbers is }15.\text{ If each number is multiplied by }2,
\displaystyle \text{what will be the new mean?}
\displaystyle \text{Answer:}
\displaystyle \text{Mean of the given numbers}=15
\displaystyle \text{If each observation is multiplied by }2,\text{ the mean is also multiplied by }2.
\displaystyle \text{New mean}=15\times2=30
\displaystyle \therefore \text{The new mean is }30.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The mean of }5\text{ numbers is }18.\text{ If one number is excluded, the mean of}
\displaystyle \text{the remaining numbers is }16.\text{ Find the excluded number.}
\displaystyle \text{Answer:}
\displaystyle \text{Total of the five numbers}=5\times18=90
\displaystyle \text{Total of the remaining four numbers}=4\times16=64
\displaystyle \text{Excluded number}=90-64=26
\displaystyle \therefore \text{The excluded number is }26.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The mean of }200\text{ items was }50.\text{ Later it was discovered that two items}
\displaystyle \text{were misread as }92\text{ and }8\text{ instead of }192\text{ and }88.\text{ Find the correct mean.}
\displaystyle \text{Answer:}
\displaystyle n=200,\qquad \text{Incorrect mean}=50
\displaystyle \text{Incorrect total}=50\times200=10000
\displaystyle \text{Correct total}=10000-92-8+192+88
\displaystyle =10180
\displaystyle \text{Correct mean}=\frac{10180}{200}=50.9
\displaystyle \therefore \text{The correct mean is }50.9.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the values of }n\text{ and }\overline{x}\text{ in each of the following cases.}

\displaystyle \text{(i) }\sum_{i=1}^{n}(x_i-12)=-10\text{ and }\sum_{i=1}^{n}(x_i-3)=62
\displaystyle \text{Answer:}
\displaystyle \sum x_i-12n=-10\qquad\ldots\text{(1)}
\displaystyle \sum x_i-3n=62\qquad\ldots\text{(2)}
\displaystyle \text{Subtracting (2) from (1),}
\displaystyle -9n=-72
\displaystyle n=8
\displaystyle \text{Substituting }n=8\text{ in (1),}
\displaystyle \sum x_i=12(8)-10=86
\displaystyle \overline{x}=\frac{\sum x_i}{n}=\frac{86}{8}=10.75
\displaystyle \therefore n=8\text{ and }\overline{x}=10.75.

\displaystyle \text{(ii) }\sum_{i=1}^{n}(x_i-10)=30\text{ and }\sum_{i=1}^{n}(x_i-6)=150
\displaystyle \text{Answer:}
\displaystyle \sum x_i-10n=30\qquad\ldots\text{(1)}
\displaystyle \sum x_i-6n=150\qquad\ldots\text{(2)}
\displaystyle \text{Subtracting (2) from (1),}
\displaystyle -4n=-120
\displaystyle n=30
\displaystyle \text{Substituting }n=30\text{ in (1),}
\displaystyle \sum x_i=10(30)+30=330
\displaystyle \overline{x}=\frac{\sum x_i}{n}=\frac{330}{30}=11
\displaystyle \therefore n=30\text{ and }\overline{x}=11.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The sums of the deviations of a set of }n\text{ values }x_1,x_2,\ldots,x_n
\displaystyle \text{measured from }15\text{ and }-3\text{ are }-90\text{ and }54\text{ respectively. Find }n\text{ and the mean.}
\displaystyle \text{Answer:}
\displaystyle \sum_{i=1}^{n}(x_i-15)=-90
\displaystyle \Rightarrow \sum x_i-15n=-90\qquad\ldots\text{(1)}
\displaystyle \sum_{i=1}^{n}(x_i+3)=54
\displaystyle \Rightarrow \sum x_i+3n=54\qquad\ldots\text{(2)}
\displaystyle \text{Subtracting (2) from (1),}
\displaystyle -18n=-144
\displaystyle n=8
\displaystyle \text{Substituting }n=8\text{ in (1),}
\displaystyle \sum x_i=15(8)-90=30
\displaystyle \overline{x}=\frac{\sum x_i}{n}=\frac{30}{8}=3.75
\displaystyle \therefore n=8\text{ and the mean is }3.75.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Find the sum of the deviations of the variate values }3,4,6,7,8,14
\displaystyle \text{from their mean.}
\displaystyle \text{Answer:}
\displaystyle n=6
\displaystyle \text{Mean }(\overline{x})=\frac{3+4+6+7+8+14}{6}=\frac{42}{6}=7
\displaystyle \sum_{i=1}^{6}(x_i-\overline{x})
\displaystyle =(3-7)+(4-7)+(6-7)+(7-7)+(8-7)+(14-7)
\displaystyle =-4-3-1+0+1+7=0
\displaystyle \therefore \text{The sum of the deviations from the mean is }0.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }\overline{x}\text{ is the mean of the ten natural numbers }x_1,x_2,\ldots,x_{10},
\displaystyle \text{show that }(x_1-\overline{x})+(x_2-\overline{x})+\cdots+(x_{10}-\overline{x})=0.
\displaystyle \text{Answer:}
\displaystyle \overline{x}=\frac{x_1+x_2+\cdots+x_{10}}{10}
\displaystyle \Rightarrow 10\overline{x}=x_1+x_2+\cdots+x_{10}
\displaystyle \Rightarrow x_1+x_2+\cdots+x_{10}-10\overline{x}=0
\displaystyle \Rightarrow (x_1-\overline{x})+(x_2-\overline{x})+\cdots+(x_{10}-\overline{x})=0
\displaystyle \therefore \sum_{i=1}^{10}(x_i-\overline{x})=0.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the mean of the observations }45,52,60,x,69,70,26,81,94\text{ is }68,
\displaystyle \text{find the value of }x.
\displaystyle \text{Answer:}
\displaystyle n=9,\qquad \text{Mean}=68
\displaystyle \text{Sum of all observations}=68\times9=612
\displaystyle \text{Sum of the known observations}=45+52+60+69+70+26+81+94=497
\displaystyle x=612-497=115
\displaystyle \therefore x=115.
\displaystyle \\

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