NCERT Solutions · Class 11 Mathematics Statistics

Mean and variance of raw data. \[\bar{x} = \frac{1}{n}\sum x_i, \qquad \sigma^2 = \frac{1}{n}\sum (x_i - \bar{x})^2 \]Here \(\displaystyle n = 8\) and \(\displaystyle 6+7+10+12+13+4+8+12 = 72\), so \[\bar{x} = \frac{72}{8} = 9 \]Deviations \(\displaystyle x_i - 9\) and their squares:

\(\displaystyle x_i\)	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 10$	$\displaystyle 12$	$\displaystyle 13$	$\displaystyle 4$	$\displaystyle 8$	$\displaystyle 12$
\(\displaystyle x_i-\bar{x}\)	\(\displaystyle -3\)	\(\displaystyle -2\)	\(\displaystyle 1\)	\(\displaystyle 3\)	\(\displaystyle 4\)	\(\displaystyle -5\)	\(\displaystyle -1\)	\(\displaystyle 3\)
\(\displaystyle (x_i-\bar{x})^2\)	$\displaystyle 9$	$\displaystyle 4$	$\displaystyle 1$	$\displaystyle 9$	$\displaystyle 16$	$\displaystyle 25$	$\displaystyle 1$	$\displaystyle 9$

\[\sum (x_i-\bar{x})^2 = 74 \qquad \Rightarrow \qquad \sigma^2 = \frac{74}{8} = 9.25 \]Mean \(\displaystyle = 9\) and variance \(\displaystyle = 9.25\).

Standard sums for the first \(\displaystyle n\) natural numbers. Use \[\sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \qquad \sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}, \qquad \sigma^{2} = \frac{\sum x_i^{2}}{n} - \bar{x}^{2} \]Mean. \[\bar{x} = \frac{1}{n}\cdot\frac{n(n+1)}{2} = \frac{n+1}{2} \]Variance. \[\frac{\sum x_i^{2}}{n} = \frac{(n+1)(2n+1)}{6} \] \[\sigma^{2} = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^{2} = (n+1)\left[\frac{2(2n+1) - 3(n+1)}{12}\right] \] \[= \frac{(n+1)(4n+2-3n-3)}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^{2}-1}{12} \]Check with \(\displaystyle n = 5\): the data \(\displaystyle 1,2,3,4,5\) has mean \(\displaystyle 3\) and \(\displaystyle \sum(x_i-3)^2 = 4+1+0+1+4 = 10\), so \(\displaystyle \sigma^2 = 2\) — and \(\displaystyle \dfrac{5^2-1}{12} = 2\). ✓Mean \(\displaystyle = \dfrac{n+1}{2}\) and variance \(\displaystyle = \dfrac{n^{2}-1}{12}\).

Scale the first $\displaystyle 10$ natural numbers. The data is \(\displaystyle 3, 6, 9, \dots, 30\), i.e. \(\displaystyle x_i = 3i\) for \(\displaystyle i = 1, 2, \dots, 10\). Multiplying every observation by \(\displaystyle 3\) multiplies the mean by \(\displaystyle 3\) and the variance by \(\displaystyle 3^{2}\).For the first \(\displaystyle n\) natural numbers, mean \(\displaystyle = \dfrac{n+1}{2}\) and variance \(\displaystyle = \dfrac{n^{2}-1}{12}\); with \(\displaystyle n = 10\) these are \(\displaystyle 5.5\) and \(\displaystyle \dfrac{99}{12} = 8.25\).Mean. \[\bar{x} = 3 \times 5.5 = 16.5 \]Variance. \[\sigma^{2} = 3^{2} \times 8.25 = 9 \times 8.25 = 74.25 \]As a direct check: \(\displaystyle \sum x_i = 3(1+2+\cdots+10) = 3 \times 55 = 165\), so \(\displaystyle \bar{x} = \dfrac{165}{10} = 16.5\), and \(\displaystyle \sum (x_i - 16.5)^2 = 742.5\), giving \(\displaystyle \sigma^2 = \dfrac{742.5}{10} = 74.25\). ✓Mean \(\displaystyle = 16.5\) and variance \(\displaystyle = 74.25\).

Mean and variance of a discrete frequency distribution. With \(\displaystyle N = \sum f_i\), \[\bar{x} = \frac{\sum f_i x_i}{N}, \qquad \sigma^{2} = \frac{\sum f_i (x_i - \bar{x})^{2}}{N} \]Here \(\displaystyle \sum f_i x_i = 760 \) and \(\displaystyle N = 40 \), so \[\bar{x} = \frac{760}{40} = 19 \]

\(\displaystyle x_i\)	\(\displaystyle f_i\)	\(\displaystyle f_i x_i\)	\(\displaystyle x_i-\bar{x}\)	\(\displaystyle (x_i-\bar{x})^2\)	\(\displaystyle f_i(x_i-\bar{x})^2\)
$\displaystyle 6$	$\displaystyle 2$	$\displaystyle 12$	\(\displaystyle -13\)	$\displaystyle 169$	$\displaystyle 338$
$\displaystyle 10$	$\displaystyle 4$	$\displaystyle 40$	\(\displaystyle -9\)	$\displaystyle 81$	$\displaystyle 324$
$\displaystyle 14$	$\displaystyle 7$	$\displaystyle 98$	\(\displaystyle -5\)	$\displaystyle 25$	$\displaystyle 175$
$\displaystyle 18$	$\displaystyle 12$	$\displaystyle 216$	\(\displaystyle -1\)	$\displaystyle 1$	$\displaystyle 12$
$\displaystyle 24$	$\displaystyle 8$	$\displaystyle 192$	$\displaystyle 5$	$\displaystyle 25$	$\displaystyle 200$
$\displaystyle 28$	$\displaystyle 4$	$\displaystyle 112$	$\displaystyle 9$	$\displaystyle 81$	$\displaystyle 324$
$\displaystyle 30$	$\displaystyle 3$	$\displaystyle 90$	$\displaystyle 11$	$\displaystyle 121$	$\displaystyle 363$
Total	$\displaystyle 40$	$\displaystyle 760$			$\displaystyle 1736$

\[\sigma^{2} = \frac{1736}{40} = 43.4 \]Mean \(\displaystyle = 19\) and variance \(\displaystyle = 43.4\).

Mean and variance of a discrete frequency distribution. With \(\displaystyle N = \sum f_i\), \[\bar{x} = \frac{\sum f_i x_i}{N}, \qquad \sigma^{2} = \frac{\sum f_i (x_i - \bar{x})^{2}}{N} \]Here \(\displaystyle \sum f_i x_i = 2200 \) and \(\displaystyle N = 22 \), so \[\bar{x} = \frac{2200}{22} = 100 \]

\(\displaystyle x_i\)	\(\displaystyle f_i\)	\(\displaystyle f_i x_i\)	\(\displaystyle x_i-\bar{x}\)	\(\displaystyle (x_i-\bar{x})^2\)	\(\displaystyle f_i(x_i-\bar{x})^2\)
$\displaystyle 92$	$\displaystyle 3$	$\displaystyle 276$	\(\displaystyle -8\)	$\displaystyle 64$	$\displaystyle 192$
$\displaystyle 93$	$\displaystyle 2$	$\displaystyle 186$	\(\displaystyle -7\)	$\displaystyle 49$	$\displaystyle 98$
$\displaystyle 97$	$\displaystyle 3$	$\displaystyle 291$	\(\displaystyle -3\)	$\displaystyle 9$	$\displaystyle 27$
$\displaystyle 98$	$\displaystyle 2$	$\displaystyle 196$	\(\displaystyle -2\)	$\displaystyle 4$	$\displaystyle 8$
$\displaystyle 102$	$\displaystyle 6$	$\displaystyle 612$	$\displaystyle 2$	$\displaystyle 4$	$\displaystyle 24$
$\displaystyle 104$	$\displaystyle 3$	$\displaystyle 312$	$\displaystyle 4$	$\displaystyle 16$	$\displaystyle 48$
$\displaystyle 109$	$\displaystyle 3$	$\displaystyle 327$	$\displaystyle 9$	$\displaystyle 81$	$\displaystyle 243$
Total	$\displaystyle 22$	$\displaystyle 2200$			$\displaystyle 640$

\[\sigma^{2} = \frac{640}{22} = \frac{320}{11} \approx 29.09 \]Mean \(\displaystyle = 100\) and variance \(\displaystyle = \dfrac{320}{11} \approx 29.09\).

Short-cut (assumed mean) method. Pick a convenient assumed mean \(\displaystyle A\) from among the \(\displaystyle x_i\), put \(\displaystyle d_i = x_i - A\), and use \[\bar{x} = A + \frac{\sum f_i d_i}{N}, \qquad \sigma^{2} = \frac{\sum f_i d_i^{2}}{N} - \left(\frac{\sum f_i d_i}{N}\right)^{2}, \qquad \sigma = \sqrt{\sigma^{2}} \] This works because shifting every observation by a constant shifts the mean by that constant and leaves the spread untouched.Take \(\displaystyle A = 64\), the value sitting near the middle of the heaviest frequencies.

\(\displaystyle x_i\)	\(\displaystyle f_i\)	\(\displaystyle d_i = x_i-64\)	\(\displaystyle f_i d_i\)	\(\displaystyle f_i d_i^{2}\)
$\displaystyle 60$	$\displaystyle 2$	\(\displaystyle -4\)	\(\displaystyle -8\)	$\displaystyle 32$
$\displaystyle 61$	$\displaystyle 1$	\(\displaystyle -3\)	\(\displaystyle -3\)	$\displaystyle 9$
$\displaystyle 62$	$\displaystyle 12$	\(\displaystyle -2\)	\(\displaystyle -24\)	$\displaystyle 48$
$\displaystyle 63$	$\displaystyle 29$	\(\displaystyle -1\)	\(\displaystyle -29\)	$\displaystyle 29$
$\displaystyle 64$	$\displaystyle 25$	$\displaystyle 0$	$\displaystyle 0$	$\displaystyle 0$
$\displaystyle 65$	$\displaystyle 12$	$\displaystyle 1$	$\displaystyle 12$	$\displaystyle 12$
$\displaystyle 66$	$\displaystyle 10$	$\displaystyle 2$	$\displaystyle 20$	$\displaystyle 40$
$\displaystyle 67$	$\displaystyle 4$	$\displaystyle 3$	$\displaystyle 12$	$\displaystyle 36$
$\displaystyle 68$	$\displaystyle 5$	$\displaystyle 4$	$\displaystyle 20$	$\displaystyle 80$
Total	$\displaystyle 100$		$\displaystyle 0$	$\displaystyle 286$

\[\bar{x} = 64 + \frac{0}{100} = 64 \] \[\sigma^{2} = \frac{286}{100} - \left(\frac{0}{100}\right)^{2} = 2.86 \] \[\sigma = \sqrt{2.86} \approx 1.69 \]Mean \(\displaystyle = 64\), variance \(\displaystyle = 2.86\) and standard deviation \(\displaystyle \approx 1.69\).

Direct method on class marks. A grouped class is represented by its class mark \(\displaystyle x_i=\dfrac{\text{lower limit}+\text{upper limit}}{2} \), and then \[\bar{x}=\frac{\sum f_i x_i}{N},\qquad \sigma^{2}=\frac{1}{N}\sum f_i\left(x_i-\bar{x}\right)^{2},\qquad N=\sum f_i. \]

Class	\(\displaystyle x_i\)	\(\displaystyle f_i\)	\(\displaystyle f_i x_i\)	\(\displaystyle x_i-\bar{x}\)	\(\displaystyle f_i\left(x_i-\bar{x}\right)^{2}\)
$\displaystyle 0$–$\displaystyle 30$	$\displaystyle 15$	$\displaystyle 2$	$\displaystyle 30$	\(\displaystyle -92\)	$\displaystyle 16928$
$\displaystyle 30$–$\displaystyle 60$	$\displaystyle 45$	$\displaystyle 3$	$\displaystyle 135$	\(\displaystyle -62\)	$\displaystyle 11532$
$\displaystyle 60$–$\displaystyle 90$	$\displaystyle 75$	$\displaystyle 5$	$\displaystyle 375$	\(\displaystyle -32\)	$\displaystyle 5120$
$\displaystyle 90$–$\displaystyle 120$	$\displaystyle 105$	$\displaystyle 10$	$\displaystyle 1050$	\(\displaystyle -2\)	$\displaystyle 40$
$\displaystyle 120$–$\displaystyle 150$	$\displaystyle 135$	$\displaystyle 3$	$\displaystyle 405$	$\displaystyle 28$	$\displaystyle 2352$
$\displaystyle 150$–$\displaystyle 180$	$\displaystyle 165$	$\displaystyle 5$	$\displaystyle 825$	$\displaystyle 58$	$\displaystyle 16820$
$\displaystyle 180$–$\displaystyle 210$	$\displaystyle 195$	$\displaystyle 2$	$\displaystyle 390$	$\displaystyle 88$	$\displaystyle 15488$
Total		$\displaystyle 30$	$\displaystyle 3210$		$\displaystyle 68280$

The mean must be found first, since the deviation column depends on it: \[\bar{x}=\frac{3210}{30}=107. \]Then \[\sigma^{2}=\frac{68280}{30}=2276. \]Mean \(\displaystyle =107\), Variance \(\displaystyle =2276\).

Direct method on class marks. Represent each class by its class mark \(\displaystyle x_i\) and use \[\bar{x}=\frac{\sum f_i x_i}{N},\qquad \sigma^{2}=\frac{1}{N}\sum f_i\left(x_i-\bar{x}\right)^{2}. \]

Class	\(\displaystyle x_i\)	\(\displaystyle f_i\)	\(\displaystyle f_i x_i\)	\(\displaystyle x_i-\bar{x}\)	\(\displaystyle f_i\left(x_i-\bar{x}\right)^{2}\)
$\displaystyle 0$–$\displaystyle 10$	$\displaystyle 5$	$\displaystyle 5$	$\displaystyle 25$	\(\displaystyle -22\)	$\displaystyle 2420$
$\displaystyle 10$–$\displaystyle 20$	$\displaystyle 15$	$\displaystyle 8$	$\displaystyle 120$	\(\displaystyle -12\)	$\displaystyle 1152$
$\displaystyle 20$–$\displaystyle 30$	$\displaystyle 25$	$\displaystyle 15$	$\displaystyle 375$	\(\displaystyle -2\)	$\displaystyle 60$
$\displaystyle 30$–$\displaystyle 40$	$\displaystyle 35$	$\displaystyle 16$	$\displaystyle 560$	$\displaystyle 8$	$\displaystyle 1024$
$\displaystyle 40$–$\displaystyle 50$	$\displaystyle 45$	$\displaystyle 6$	$\displaystyle 270$	$\displaystyle 18$	$\displaystyle 1944$
Total		$\displaystyle 50$	$\displaystyle 1350$		$\displaystyle 6600$

\[\bar{x}=\frac{1350}{50}=27,\qquad \sigma^{2}=\frac{6600}{50}=132. \]Mean \(\displaystyle =27\), Variance \(\displaystyle =132\).

Short-cut (step-deviation) method. All classes have the same width \(\displaystyle h=5\), so put \(\displaystyle y_i=\dfrac{x_i-A}{h}\) with assumed mean \(\displaystyle A=92.5\) (the class mark of $\displaystyle 90$–$\displaystyle 95$, near the middle of the data). Then \[\bar{x}=A+h\left(\frac{\sum f_i y_i}{N}\right),\qquad \sigma^{2}=h^{2}\left[\frac{\sum f_i y_i^{2}}{N}-\left(\frac{\sum f_i y_i}{N}\right)^{2}\right]. \]

Height (cm)	\(\displaystyle x_i\)	\(\displaystyle f_i\)	\(\displaystyle y_i=\dfrac{x_i-92.5}{5}\)	\(\displaystyle f_i y_i\)	\(\displaystyle f_i y_i^{2}\)
$\displaystyle 70$–$\displaystyle 75$	$\displaystyle 72.5$	$\displaystyle 3$	\(\displaystyle -4\)	\(\displaystyle -12\)	$\displaystyle 48$
$\displaystyle 75$–$\displaystyle 80$	$\displaystyle 77.5$	$\displaystyle 4$	\(\displaystyle -3\)	\(\displaystyle -12\)	$\displaystyle 36$
$\displaystyle 80$–$\displaystyle 85$	$\displaystyle 82.5$	$\displaystyle 7$	\(\displaystyle -2\)	\(\displaystyle -14\)	$\displaystyle 28$
$\displaystyle 85$–$\displaystyle 90$	$\displaystyle 87.5$	$\displaystyle 7$	\(\displaystyle -1\)	\(\displaystyle -7\)	$\displaystyle 7$
$\displaystyle 90$–$\displaystyle 95$	$\displaystyle 92.5$	$\displaystyle 15$	$\displaystyle 0$	$\displaystyle 0$	$\displaystyle 0$
$\displaystyle 95$–$\displaystyle 100$	$\displaystyle 97.5$	$\displaystyle 9$	$\displaystyle 1$	$\displaystyle 9$	$\displaystyle 9$
$\displaystyle 100$–$\displaystyle 105$	$\displaystyle 102.5$	$\displaystyle 6$	$\displaystyle 2$	$\displaystyle 12$	$\displaystyle 24$
$\displaystyle 105$–$\displaystyle 110$	$\displaystyle 107.5$	$\displaystyle 6$	$\displaystyle 3$	$\displaystyle 18$	$\displaystyle 54$
$\displaystyle 110$–$\displaystyle 115$	$\displaystyle 112.5$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 12$	$\displaystyle 48$
Total		$\displaystyle 60$		$\displaystyle 6$	$\displaystyle 254$

Mean: \[\bar{x}=92.5+5\times\frac{6}{60}=92.5+0.5=93. \]Variance: \[\sigma^{2}=5^{2}\left[\frac{254}{60}-\left(\frac{6}{60}\right)^{2}\right]=25\left[\frac{254}{60}-\frac{1}{100}\right]=25\times\frac{1267}{300}=\frac{1267}{12}\approx 105.58. \]Standard deviation: \[\sigma=\sqrt{105.583\ldots}\approx 10.28. \]Mean \(\displaystyle =93\) cm, Variance \(\displaystyle \approx 105.58\), Standard deviation \(\displaystyle \approx 10.28\) cm.

Short-cut method after making the classes continuous. The given classes $\displaystyle 33$–$\displaystyle 36$, $\displaystyle 37$–$\displaystyle 40$, \(\displaystyle \ldots\) are inclusive: there is a gap of \(\displaystyle 1\) mm between one upper limit and the next lower limit. Subtract \(\displaystyle 0.5\) from each lower limit and add \(\displaystyle 0.5\) to each upper limit (as the hint says) to get the continuous classes $\displaystyle 32.5$–$\displaystyle 36.5$, $\displaystyle 36.5$–$\displaystyle 40.5$, $\displaystyle 40.5$–$\displaystyle 44.5$, $\displaystyle 44.5$–$\displaystyle 48.5$, $\displaystyle 48.5$–$\displaystyle 52.5$, each of width \(\displaystyle h=4\).With \(\displaystyle A=42.5\) and \(\displaystyle y_i=\dfrac{x_i-42.5}{4}\), \[\bar{x}=A+h\left(\frac{\sum f_i y_i}{N}\right),\qquad \sigma^{2}=h^{2}\left[\frac{\sum f_i y_i^{2}}{N}-\left(\frac{\sum f_i y_i}{N}\right)^{2}\right]. \]

Class (continuous)	\(\displaystyle x_i\)	\(\displaystyle f_i\)	\(\displaystyle y_i\)	\(\displaystyle f_i y_i\)	\(\displaystyle f_i y_i^{2}\)
$\displaystyle 32.5$–$\displaystyle 36.5$	$\displaystyle 34.5$	$\displaystyle 15$	\(\displaystyle -2\)	\(\displaystyle -30\)	$\displaystyle 60$
$\displaystyle 36.5$–$\displaystyle 40.5$	$\displaystyle 38.5$	$\displaystyle 17$	\(\displaystyle -1\)	\(\displaystyle -17\)	$\displaystyle 17$
$\displaystyle 40.5$–$\displaystyle 44.5$	$\displaystyle 42.5$	$\displaystyle 21$	$\displaystyle 0$	$\displaystyle 0$	$\displaystyle 0$
$\displaystyle 44.5$–$\displaystyle 48.5$	$\displaystyle 46.5$	$\displaystyle 22$	$\displaystyle 1$	$\displaystyle 22$	$\displaystyle 22$
$\displaystyle 48.5$–$\displaystyle 52.5$	$\displaystyle 50.5$	$\displaystyle 25$	$\displaystyle 2$	$\displaystyle 50$	$\displaystyle 100$
Total		$\displaystyle 100$		$\displaystyle 25$	$\displaystyle 199$

Mean: \[\bar{x}=42.5+4\times\frac{25}{100}=42.5+1=43.5. \]Variance: \[\sigma^{2}=4^{2}\left[\frac{199}{100}-\left(\frac{25}{100}\right)^{2}\right]=16\left[1.99-0.0625\right]=16\times 1.9275=30.84. \]Standard deviation: \[\sigma=\sqrt{30.84}\approx 5.55. \]Note that the correction leaves every class mark unchanged \(\displaystyle \left(\tfrac{33+36}{2}=34.5=\tfrac{32.5+36.5}{2}\right)\); what it fixes is the class width, which is \(\displaystyle 4\) and not \(\displaystyle 3\).Mean diameter \(\displaystyle =43.5\) mm, Standard deviation \(\displaystyle \approx 5.55\) mm (variance \(\displaystyle =30.84\)).