NCERT Solutions · Class 11 Mathematics Statistics

Use the two totals \(\displaystyle \sum x_i\) and \(\displaystyle \sum x_i^{2}\). For \(\displaystyle n\) observations, \[\bar{x}=\frac{\sum x_i}{n},\qquad \sigma^{2}=\frac{\sum x_i^{2}}{n}-\bar{x}^{2}. \]Let the two unknown observations be \(\displaystyle a\) and \(\displaystyle b\), with \(\displaystyle n=8\), \(\displaystyle \bar{x}=9\), \(\displaystyle \sigma^{2}=9.25\).Sum. \(\displaystyle \sum x_i=8\times 9=72\). The six known observations add to \[6+7+10+12+12+13=60, \] so \[a+b=72-60=12. \tag{1} \]Sum of squares. From \(\displaystyle \sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\bar{x}^{2}\), \[\sum x_i^{2}=8\left(9.25+81\right)=8\times 90.25=722. \] The six known squares add to \[36+49+100+144+144+169=642, \] so \[a^{2}+b^{2}=722-642=80. \tag{2} \]Solve ($\displaystyle 1$) and ($\displaystyle 2$). From \(\displaystyle (a+b)^{2}=a^{2}+b^{2}+2ab\), \[144=80+2ab\ \Rightarrow\ ab=32. \] So \(\displaystyle a\) and \(\displaystyle b\) are the roots of \(\displaystyle t^{2}-12t+32=0\), i.e. \(\displaystyle (t-4)(t-8)=0\).Check: with $\displaystyle 4$ and $\displaystyle 8$ the eight values sum to \(\displaystyle 72\) (mean $\displaystyle 9$) and their squares sum to \(\displaystyle 642+16+64=722\), giving \(\displaystyle \dfrac{722}{8}-81=9.25\). ✓The remaining two observations are $\displaystyle 4$ and 8.

Use the two totals \(\displaystyle \sum x_i\) and \(\displaystyle \sum x_i^{2}\). Here \(\displaystyle n=7\), \(\displaystyle \bar{x}=8\), \(\displaystyle \sigma^{2}=16\). Let the missing observations be \(\displaystyle a\) and \(\displaystyle b\).Sum. \(\displaystyle \sum x_i=7\times 8=56\), and \(\displaystyle 2+4+10+12+14=42\), so \[a+b=56-42=14. \tag{1} \]Sum of squares. From \(\displaystyle \sigma^{2}=\dfrac{\sum x_i^{2}}{n}-\bar{x}^{2}\), \[\sum x_i^{2}=7\left(16+64\right)=560, \] and \(\displaystyle 4+16+100+144+196=460\), so \[a^{2}+b^{2}=560-460=100. \tag{2} \]Solve. \(\displaystyle (a+b)^{2}=a^{2}+b^{2}+2ab\) gives \(\displaystyle 196=100+2ab\), so \(\displaystyle ab=48\). Hence \(\displaystyle a,b\) are the roots of \[t^{2}-14t+48=0\ \Rightarrow\ (t-6)(t-8)=0. \]Check: \(\displaystyle 2+4+10+12+14+6+8=56\) so the mean is \(\displaystyle 8\); \(\displaystyle 460+36+64=560\) and \(\displaystyle \dfrac{560}{7}-64=80-64=16\). ✓The remaining two observations are $\displaystyle 6$ and 8.

Effect of multiplying every observation by a constant. If each \(\displaystyle x_i\) is replaced by \(\displaystyle k x_i\), then \[\overline{kx}=\frac{1}{n}\sum k x_i=k\bar{x}, \] \[\text{Var}(kx)=\frac{1}{n}\sum\left(kx_i-k\bar{x}\right)^{2}=k^{2}\cdot\frac{1}{n}\sum\left(x_i-\bar{x}\right)^{2}=k^{2}\sigma^{2}, \] so the standard deviation becomes \(\displaystyle |k|\sigma\).Here \(\displaystyle \bar{x}=8\), \(\displaystyle \sigma=4\) and \(\displaystyle k=3\): \[\text{new mean}=3\times 8=24,\qquad \text{new S.D.}=|3|\times 4=12. \](The number of observations, six, is not needed — the result holds for any \(\displaystyle n\).)New mean \(\displaystyle =24\), new standard deviation \(\displaystyle =12\).

Proof from the definitions of mean and variance.Let \(\displaystyle y_i=a x_i\) for \(\displaystyle i=1,2,\ldots,n\), where \(\displaystyle a\neq 0\). We are given \[\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i,\qquad \sigma^{2}=\frac{1}{n}\sum_{i=1}^{n}\left(x_i-\bar{x}\right)^{2}. \]Mean of the new observations. \[\bar{y}=\frac{1}{n}\sum_{i=1}^{n}y_i=\frac{1}{n}\sum_{i=1}^{n}a x_i=a\cdot\frac{1}{n}\sum_{i=1}^{n}x_i=a\bar{x}, \] using that the constant \(\displaystyle a\) can be taken outside the summation.Variance of the new observations. By definition, and substituting \(\displaystyle \bar{y}=a\bar{x}\), \[\text{Var}(y)=\frac{1}{n}\sum_{i=1}^{n}\left(y_i-\bar{y}\right)^{2}=\frac{1}{n}\sum_{i=1}^{n}\left(a x_i-a\bar{x}\right)^{2}=\frac{1}{n}\sum_{i=1}^{n}a^{2}\left(x_i-\bar{x}\right)^{2}. \] Again taking the constant \(\displaystyle a^{2}\) outside, \[\text{Var}(y)=a^{2}\cdot\frac{1}{n}\sum_{i=1}^{n}\left(x_i-\bar{x}\right)^{2}=a^{2}\sigma^{2}. \]Hence the observations \(\displaystyle a x_1,a x_2,\ldots,a x_n\) have mean \(\displaystyle a\bar{x}\) and variance \(\displaystyle a^{2}\sigma^{2}\). \(\displaystyle \blacksquare\)(The condition \(\displaystyle a\neq 0\) only rules out the degenerate case in which every new observation collapses to \(\displaystyle 0\); the algebra above is valid for every real \(\displaystyle a\).)

Correct the two totals \(\displaystyle \sum x_i\) and \(\displaystyle \sum x_i^{2}\), then re-apply the formulas. For the original $\displaystyle 20$ observations, \(\displaystyle \bar{x}=10\) and \(\displaystyle \sigma=2\), so \[\sum x_i=20\times 10=200, \] \[\sum x_i^{2}=n\left(\sigma^{2}+\bar{x}^{2}\right)=20\left(4+100\right)=2080. \](i) The wrong item $\displaystyle 8$ is omitted. Remove \(\displaystyle 8\) from the first total and \(\displaystyle 8^{2}=64\) from the second, and drop \(\displaystyle n\) to $\displaystyle 19$: \[\sum x_i=200-8=192,\qquad \sum x_i^{2}=2080-64=2016. \] \[\bar{x}=\frac{192}{19}\approx 10.11\ (=10.1\text{ to one decimal place}). \] \[\sigma^{2}=\frac{2016}{19}-\left(\frac{192}{19}\right)^{2}=\frac{19\times 2016-36864}{361}=\frac{38304-36864}{361}=\frac{1440}{361}\approx 3.989, \] \[\sigma=\frac{\sqrt{1440}}{19}=\frac{12\sqrt{10}}{19}\approx 1.997. \](ii) The wrong item $\displaystyle 8$ is replaced by 12. Now \(\displaystyle n\) stays $\displaystyle 20$: \[\sum x_i=200-8+12=204,\qquad \sum x_i^{2}=2080-64+144=2160. \] \[\bar{x}=\frac{204}{20}=10.2, \] \[\sigma^{2}=\frac{2160}{20}-\left(10.2\right)^{2}=108-104.04=3.96,\qquad \sigma=\sqrt{3.96}=\frac{3\sqrt{11}}{5}\approx 1.99. \](i) Mean \(\displaystyle \approx 10.1\), S.D. \(\displaystyle \approx 1.997\). (ii) Mean \(\displaystyle =10.2\), S.D. \(\displaystyle \approx 1.99\).

Correct the two totals \(\displaystyle \sum x_i\) and \(\displaystyle \sum x_i^{2}\), then re-apply the formulas. For the original $\displaystyle 100$ observations, \(\displaystyle \bar{x}=20\) and \(\displaystyle \sigma=3\), so \[\sum x_i=100\times 20=2000, \] \[\sum x_i^{2}=n\left(\sigma^{2}+\bar{x}^{2}\right)=100\left(9+400\right)=40900. \]Drop the three incorrect entries $\displaystyle 21$, $\displaystyle 21$, 18. \[21+21+18=60,\qquad 21^{2}+21^{2}+18^{2}=441+441+324=1206. \] So for the remaining \(\displaystyle n=97\) observations, \[\sum x_i=2000-60=1940,\qquad \sum x_i^{2}=40900-1206=39694. \]New mean. \[\bar{x}=\frac{1940}{97}=20. \] (The three discarded values averaged \(\displaystyle 20\) themselves, so the mean is unchanged.)New variance and standard deviation. \[\sigma^{2}=\frac{39694}{97}-\left(20\right)^{2}=\frac{39694-38800}{97}=\frac{894}{97}\approx 9.2165, \] \[\sigma=\sqrt{9.2165\ldots}\approx 3.036. \]Mean \(\displaystyle =20\), Standard deviation \(\displaystyle \approx 3.04\) (more precisely \(\displaystyle 3.036\); variance \(\displaystyle \approx 9.217\)).