From the prices of shares X and Y below, find out which is more stable in value:
Question: From the prices of shares X and Y below, find out which is more stable in value: Solution: The prices of the shares X are 35, 54, 52, 53, 56, 58, 52, 50, 51, 49 Here, the number of observations, N = 10 $\therefore$ Mean, $\bar{x}=\frac{1}{N} \sum_{i=1}^{10} x_{i}=\frac{1}{10} \times 510=51$ The following table is obtained corresponding to shares X. Variance $\left(\sigma_{1}^{2}\right)=\frac{1}{N} \sum_{i=1}^{10}(x i-\bar{x})^{2}=\frac{1}{10} \times 350=35$ $\therefore S \tan$ dard dev...
Read More →From the data given below state which group is more variable, A or B?
Question: From the data given below state which group is more variable, A or B? Solution: Firstly, the standard deviation of group A is calculated as follows. Here,h= 10, N = 150, A = 45 Mean $=\mathrm{A}+\frac{\sum_{\mathrm{i}=1}^{7} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}} \times \mathrm{h}=45+\frac{(-6) \times 10}{150}=45-0.4=44.6$ $\sigma_{1}^{2}=\frac{h^{2}}{N^{2}}\left(N \sum_{i=1}^{7} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{7} f_{i} y_{i}\right)^{2}\right)$ $=\frac{100}{22500}\left(150 \times 342-(...
Read More →If the radius of a sphere is measured as 9 m with an error of 0.03 m,
Question: If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating in surface area. Solution: Letrbe the radius of the sphere and Δrbe the error in measuring the radius. Then, $r=9 \mathrm{~m}$ and $\Delta r=0.03 \mathrm{~m}$ Now, the surface area of the sphere (S)is given by, $S=4 \pi r^{2}$ $\therefore \frac{d S}{d r}=8 \pi r$ $\therefore d S=\left(\frac{d S}{d r}\right) \Delta r$ $=(8 \pi r) \Delta r$ $=8 \pi(9)(0.03) \mathrm{m}^{2}$...
Read More →Prove that:
Question: Prove that: (i) $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=a b c$ (ii) $\left(\mathrm{a}^{-1}+\mathrm{b}^{-1}\right)^{-1}$ Solution: (i) To prove, $=\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=a b c$ Left hand side (LHS) = Right hand side (RHS) Considering LHS, $=\frac{a+b+c}{\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}}$ $=\frac{a+b+c}{\frac{a+b+c}{a b c}}$ = abc Therefore, LHS = RHS Hence proved (ii) To prove, $\left(\mathrm{a}^{-1}+\mathrm{b}^{-1...
Read More →If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.
Question: If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume. Solution: Letrbe the radius of the sphere and Δrbe the error in measuring the radius. Then, $r=7 \mathrm{~m}$ and $\Delta r=0.02 \mathrm{~m}$ Now, the volumeVof the sphere is given by, $V=\frac{4}{3} \pi r^{3}$ $\begin{aligned} \therefore \frac{d V}{d r} =4 \pi r^{2} \\ \therefore d V =\left(\frac{d V}{d r}\right) \Delta r \\ =\left(4 \pi r^{2}\right) \Delta r...
Read More →The diameters of circles (in mm) drawn in a design are given below:
Question: The diameters of circles (in mm) drawn in a design are given below: Solution: Here, N = 100,h= 4 Let the assumed mean, A, be 42.5. Mean, $\bar{x}=A+\frac{\sum_{i=1}^{5} f_{i} y_{i}}{N} \times h=42.5+\frac{25}{100} \times 4=43.5$ $\operatorname{Variance}\left(\sigma^{2}\right)=\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{3} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{3} f_{i} y_{i}\right)^{2}\right]$ $=\frac{16}{10000}\left[100 \times 199-(25)^{2}\right]$ $=\frac{16}{10000}[19900-625]$ $=\frac{16}{1000...
Read More →Find the mean, variance and standard deviation using short-cut method
Question: Find the mean, variance and standard deviation using short-cut method Solution: Mean, $\bar{x}=\mathrm{A}+\frac{\sum_{i=1}^{9} \mathrm{f}_{i} \mathrm{y}_{i}}{\mathrm{~N}} \times \mathrm{h}=92.5+\frac{6}{60} \times 5=92.5+0.5=93$ Variance $\left(\sigma^{2}\right)=\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{g} f_{i} y_{i}{ }^{2}-\left(\sum_{i=1}^{g} f_{i} y_{i}\right)^{2}\right]$ $=\frac{(5)^{2}}{(60)^{2}}\left[60 \times 254-(6)^{2}\right]$ $=\frac{25}{3600}(15204)=105.58$ $\therefore$ Stan d...
Read More →Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%
Question: Find the approximate change in the surface area of a cube of sidexmetres caused by decreasing the side by 1% Solution: The surface area of a cube $(S)$ of side $x$ is given by $S=6 x^{2}$. $\begin{aligned} \therefore \frac{d S}{d x} =\left(\frac{d S}{d x}\right) \Delta x \\ =(12 x) \Delta x \\ =(12 x)(0.01 x) [\text { as } 1 \% \text { of } x \text { is } 0.01 x] \\ =0.12 x^{2} \end{aligned}$ Hence, the approximate change in the surface area of the cube is $0.12 x^{2} \mathrm{~m}^{2}$....
Read More →Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.
Question: Find the approximate change in the volumeVof a cube of sidexmetres caused by increasing side by 1%. Solution: The volume of a cube $(V)$ of side $x$ is given by $V=x^{3}$. $\begin{aligned} \therefore d V =\left(\frac{d V}{d x}\right) \Delta x \\ =\left(3 x^{2}\right) \Delta x \\ =\left(3 x^{2}\right)(0.01 x) [\text { as } 1 \% \text { of } x \text { is } 0.01 x] \\ =0.03 x^{3} \end{aligned}$ Hence, the approximate change in the volume of the cube is $0.03 x^{3} \mathrm{~m}^{3}$....
Read More →Find the mean and variance for the following frequency distribution.
Question: Find the mean and variance for the following frequency distribution. Solution: Mean, $\frac{-}{x}=\mathrm{A}+\frac{\sum_{i=1}^{5} \mathrm{f}_{i} \mathrm{y}_{i}}{\mathrm{~N}} \times \mathrm{h}=25+\frac{10}{50} \times 10=25+2=27$ Variance $\left(\sigma^{2}\right)=\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{5} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{5} f_{i} y_{i}\right)^{2}\right]$ $=\frac{(10)^{2}}{(50)^{2}}\left[50 \times 68-(10)^{2}\right]$ $=\frac{1}{25}[3400-100]=\frac{3300}{25}$ $=132$...
Read More →Find the approximate value of
Question: Find the approximate value of $f(5.001)$, where $f(x)=x^{3}-7 x^{2}+15$. Solution: Let $x=5$ and $\Delta x=0.001$. Then, we have: $f(5.001)=f(x+\Delta x)=(x+\Delta x)^{3}-7(x+\Delta x)^{2}+15$ Now, $\Delta y=f(x+\Delta x)-f(x)$ $\therefore f(x+\Delta x)=f(x)+\Delta y$ $\approx f(x)+f^{\prime}(x) \cdot \Delta x \quad($ as $d x=\Delta x)$ $\begin{array}{rlr}\Rightarrow f(5.001) \approx\left(x^{3}-7 x^{2}+15\right)+\left(3 x^{2}-14 x\right) \Delta x \\ =\left[(5)^{3}-7(5)^{2}+15\right]+\l...
Read More →Find the mean and variance for the following frequency distribution.
Question: Find the mean and variance for the following frequency distribution. Solution: Mean, $\bar{x}=\mathrm{A}+\frac{\sum_{\mathrm{i}=1}^{7} \mathrm{f}_{\mathrm{i}} \mathrm{y}_{\mathrm{i}}}{\mathrm{N}} \times \mathrm{h}=105+\frac{2}{30} \times 30=105+2=107$ $\operatorname{Variance}\left(\sigma^{2}\right)=\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{7} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{7} f_{i} y_{i}\right)^{2}\right]$ $=\frac{(30)^{2}}{(30)^{2}}\left[30 \times 76-(2)^{2}\right]$ $=2280-4$ $=2276$...
Read More →Find the approximate value of
Question: Find the approximate value of $f(2.01)$, where $f(x)=4 x^{2}+5 x+2$ Solution: Let $x=2$ and $\Delta x=0.01$. Then, we have: $f(2.01)=f(x+\Delta x)=4(x+\Delta x)^{2}+5(x+\Delta x)+2$ Now, $\Delta y=f(x+\Delta x)-f(x)$ $\therefore f(x+\Delta x)=f(x)+\Delta y$ $\approx f(x)+f^{\prime}(x) \cdot \Delta x \quad($ as $d x=\Delta x)$ $\Rightarrow f(2.01) \approx\left(4 x^{2}+5 x+2\right)+(8 x+5) \Delta x$ $=\left[4(2)^{2}+5(2)+2\right]+[8(2)+5](0.01) \quad[$ as $x=2, \Delta x=0.01]$ $=(16+10+2...
Read More →Prove that:
Question: Prove that: (i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$ (ii) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}$ Solution: (i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$ Left hand side (LHS) = Right hand side (RHS) Considering LHS, $=\frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}$ $=\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}$ $=\frac{x^{b}+x^{a}}{x^{a}+x b}$ $=1$ Therefore, LHS = RHS Hence proved (ii) $\frac{1}{1+x^{b-a}+...
Read More →Find the mean and standard deviation using short-cut method.
Question: Find the mean and standard deviation using short-cut method. Solution: The data is obtained in tabular form as follows. Mean, $\frac{-}{x}=A \frac{\sum_{i=1}^{9} f_{i} y_{i}}{N} \times h=64+\frac{0}{100} \times 1=64+0=64$ Variance,$\sigma^{2}=\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{9} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{9} f_{i} y_{i}\right)^{2}\right]$ $=\frac{1}{100^{2}}[100 \times 286-0]$ $=2.86$ $\therefore$ Stan dard deviation $(\sigma)=\sqrt{2.86}=1.69$...
Read More →Find the mean and variance for the data
Question: Find the mean and variance for the data Solution: The data is obtained in tabular form as follows. Here, $N=22, \sum_{i=1}^{7} f_{i} x_{i}=2200$ $\therefore \bar{x}=\frac{1}{N} \sum_{i=1}^{7} f_{i} x_{i}=\frac{1}{22} \times 2200=100$ Variance $\left(\sigma^{2}\right)=\frac{1}{N} \sum_{i=1}^{7} \mathrm{f}_{i}\left(\mathrm{x}_{i}-\overline{\mathrm{x}}\right)^{2}=\frac{1}{22} \times 640=29.09$...
Read More →Using differentials, find the approximate value of each
Question: 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal (i) $\sqrt{25.3}$ (ii) $\sqrt{49.5}$ (iii) $\sqrt{0.6}$ (iv) $(0.009)^{\frac{1}{3}}$ (v) $(0.999)^{\frac{1}{10}}$ (vi) $(15)^{\frac{1}{4}}$\ (vii) $(26)^{\frac{1}{3}}$ (viii) $(255)^{\frac{1}{4}}$ (ix) $(82)^{\frac{1}{4}}$ (x) $(401)^{\frac{1}{2}}$ (xi) $(0.0037)^{\frac{1}{2}}$ (xii) $(26.57)^{\frac{1}{3}}$ (xiii) $(81.5)^{\frac{1}{4}}$ (xiv) $(3.968)^{\frac{3}{2}}$ (xv) $(32.15)^{\fra...
Read More →Find the mean and variance for the data
Question: Find the mean and variance for the data Solution: The data is obtained in tabular form as follows. Here, $N=40, \sum_{i=1}^{T} f_{i} x_{i}=760$ $\therefore \bar{x}=\frac{\sum_{i=1}^{7} f_{i} x_{i}}{N}=\frac{760}{40}=19$ Variance $=\left(\sigma^{2}\right)=\frac{1}{\mathrm{~N}} \sum_{i=1}^{7} \mathrm{f}_{i}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=\frac{1}{40} \times 1736=43.4$...
Read More →Prove that:
Question: Prove that: Solution: (i) To prove $\left(\frac{x^{a}}{x^{b}}\right)^{a^{2}+a b+b^{2}} \times\left(\frac{x^{b}}{x^{c}}\right)^{b^{2}+b c+c^{2}} \times\left(\frac{x^{c}}{x^{a}}\right)^{c^{2}+c a+a^{2}}=1$ Left hand side (LHS) = Right hand side (RHS) Considering LHS, $=\frac{x^{a^{3}}+a^{2} b+a b^{2}}{x^{a^{2} b}+a b^{2}+b^{3}} \times \frac{x^{b^{3}}+b^{2} c+b c^{2}}{x^{b^{2}} c+b c^{2}+c^{3}} \times \frac{x^{c^{3}}+c^{2} a+c a^{2}}{x^{c^{2}} a+c a^{2}+a^{3}}$ $=x^{a^{3}}+a^{2} b+a b^{2}...
Read More →Find the mean and variance for the first 10 multiples of 3
Question: Find the mean and variance for the first 10 multiples of 3 Solution: The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 Here, number of observations,n= 10 Mean, $\bar{x}=\frac{\sum_{i=1}^{10} x_{i}}{10}=\frac{165}{10}=16.5$ The following table is obtained. Variance $\left(\sigma^{2}\right)=\frac{1}{n} \sum_{i=1}^{10}\left(x_{i}-\bar{x}\right)^{2}=\frac{1}{10} \times 742.5=74.25$...
Read More →If a = 3 and b = - 2, find the values of:
Question: If $a=3$ and $b=-2$, find the values of: (i) $a^{a}+b^{b}$ (ii) $a^{b}+b^{a}$ (iii) $a^{b}+b^{a}$ Solution: (i) We have, $a^{a}+b^{b}$ $=3^{3}+(-2)^{-2}$ $=3^{3}+(-1 / 2)^{2}$ $=27+1 / 4$ $=109 / 4$ (ii) $a^{b}+b^{a}$ $=3^{-2}+(-2)^{3}$ $=(1 / 3)^{2}+(-2)^{3}$ $=1 / 9-8$ $=-(71 / 9)$ (iii) We have, $a^{b}+b^{a}$ $=(3+(-2))^{3(-2)}$ $=(3-2))^{-6}$ $=1^{-6}=1$...
Read More →Find the mean and variance for the first n natural numbers
Question: Find the mean and variance for the firstnnatural numbers Solution: The mean of firstnnatural numbers is calculated as follows. Mean $=\frac{\text { Sum of all observations }}{\text { Number of observations }}$ $\therefore$ Mean $=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$ Variance $\left(\sigma^{2}\right)=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}$ $=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}}\left[\mathrm{x}_{\mathrm{i}}-\left(\frac{\mathrm{n}+\mathrm{l}}{2}\rig...
Read More →Simplify the following:
Question: Simplify the following: (i) $3\left(a^{4} b^{3}\right)^{10} \times 5\left(a^{2} b^{2}\right)^{3}$ (ii) $\left(2 x^{-2} y^{3}\right)^{3}$ (iii) $\frac{\left(4 \times 10^{7}\right)\left(6 \times 10^{-5}\right)}{8 \times 10^{4}}$ (iv) $\frac{4 a b^{2}\left(-5 a b^{3}\right)}{10 a^{2} b^{2}}$ (v) $\left(\frac{x^{2} y^{2}}{a^{2} b^{3}}\right)^{n}$ (vi) $\frac{\left(a^{3 n}-9\right)^{6}}{a^{2 n-4}}$ Solution: (i) $3\left(a^{4} b^{3}\right)^{10} \times 5\left(a^{2} b^{2}\right)^{3}$ $=3\left(...
Read More →Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Question: Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12 Solution: 6, 7, 10, 12, 13, 4, 8, 12 Mean, $\bar{x}=\frac{\sum_{i=1}^{8} x_{i}}{n}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$ The following table is obtained. Variance $\left(\sigma^{2}\right)=\frac{1}{\mathrm{n}} \sum_{i=1}^{8}\left(\mathrm{x}_{i}-\overline{\mathrm{x}}\right)^{2}=\frac{1}{8} \times 74=9.25$...
Read More →Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Question: Calculate the mean deviation about median age for the age distribution of 100 persons given below: Solution: The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval. The table is formed as follows. The class interval containing the $\frac{\mathrm{N}^{\text {th }}}{2}$ or $50^{\text {th }}$ item is $35.5-40.5$. Therefore, 35.5 40.5 is the me...
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