NCERT Class 11 Mathematics — Statistics, Exercise 13.2. All 10 questions solved.

Variance replaces the modulus of the previous exercise with a square, which makes it algebraically tractable:

$$\sigma^2 = \frac{\sum f_i\left(x_i – \bar x\right)^2}{\mathrm N} = \frac{\sum f_ix_i^2}{\mathrm N} – \bar x^2$$

The second form is usually faster, since it needs only $\sum f_ix_i$ and $\sum f_ix_i^2$. Standard deviation is $\sigma = \sqrt{\sigma^2}$, in the same units as the data.

The short-cut (step-deviation) method shifts and scales the data first. With $y_i = \dfrac{x_i – \mathrm A}{h}$ for an assumed mean $\mathrm A$ and class width $h$:

$$\bar x = \mathrm A + \frac{\sum f_iy_i}{\mathrm N} \times h, \qquad \sigma^2 = \frac{h^2}{\mathrm N^2}\left[\mathrm N\sum f_iy_i^2 – \left(\sum f_iy_i\right)^2\right]$$

Key insight. Shifting every observation by a constant leaves the variance unchanged; scaling by $h$ multiplies it by $h^2$. That is the whole justification for the short-cut method — and it is also what questions 3 and 4 of the miscellaneous exercise are about. Choosing $\mathrm A$ near the middle of the data keeps the $y_i$ small and the arithmetic easy.

Find the mean and variance for each of the data in questions 1 to 5.

Question 1

$6,\ 7,\ 10,\ 12,\ 13,\ 4,\ 8,\ 12$

Solution. Eight observations summing to $72$:

$$\bar x = \frac{72}{8} = 9$$

Deviations from $9$: $-3, -2, 1, 3, 4, -5, -1, 3$. Their squares:

$$9, 4, 1, 9, 16, 25, 1, 9 \quad\text{summing to } 74$$

$$\sigma^2 = \frac{74}{8} = 9.25$$

Question 2

First $n$ natural numbers.

Solution. The data are $1, 2, \ldots, n$, and two standard sums are needed:

$$\sum_{i=1}^n i = \frac{n(n+1)}{2}, \qquad \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$

Mean:

$$\bar x = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$$

Variance, using $\sigma^2 = \dfrac{\sum x_i^2}{n} – \bar x^2$:

$$\sigma^2 = \frac{(n+1)(2n+1)}{6} – \frac{(n+1)^2}{4}$$

Taking out $\dfrac{n+1}{12}$:

$$= \frac{n+1}{12}\left[2(2n+1) – 3(n+1)\right] = \frac{n+1}{12}(4n + 2 – 3n – 3) = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$$

Question 3

First $10$ multiples of $3$.

Solution. The data are $3, 6, 9, \ldots, 30$, that is $3$ times the first ten natural numbers. Using question 2’s results with $n = 10$ and the scaling rule (multiplying by $3$ multiplies the mean by $3$ and the variance by $3^2$):

$$\bar x = 3 \times \frac{11}{2} = 16.5$$

$$\sigma^2 = 9 \times \frac{100 – 1}{12} = 9 \times \frac{99}{12} = \frac{891}{12} = 74.25$$

Checking directly: $\sum x_i = 165$, $\sum x_i^2 = 3465$, so $\sigma^2 = \tfrac{3465}{10} – 16.5^2 = 346.5 – 272.25 = 74.25$ ✓

Question 4

Solution. $\mathrm N = 40$.

$$\sum f_ix_i = 12 + 40 + 98 + 216 + 192 + 112 + 90 = 760 \quad\Longrightarrow\quad \bar x = \frac{760}{40} = 19$$

$$\sum f_ix_i^2 = 72 + 400 + 1372 + 3888 + 4608 + 3136 + 2700 = 16176$$

$$\sigma^2 = \frac{16176}{40} – 19^2 = 404.4 – 361 = 43.4$$

Question 5

Solution. $\mathrm N = 22$.

$$\sum f_ix_i = 276 + 186 + 291 + 196 + 612 + 312 + 327 = 2200 \quad\Longrightarrow\quad \bar x = \frac{2200}{22} = 100$$

The mean comes out exactly $100$, so the deviations $x_i – 100$ are small integers:

$$-8,\ -7,\ -3,\ -2,\ 2,\ 4,\ 9$$

$$\sum f_i(x_i – 100)^2 = 3(64) + 2(49) + 3(9) + 2(4) + 6(4) + 3(16) + 3(81)$$ $$= 192 + 98 + 27 + 8 + 24 + 48 + 243 = 640$$

$$\sigma^2 = \frac{640}{22} = \frac{320}{11} \approx 29.09$$

Question 6

Find the mean and standard deviation using the short-cut method.

Solution. Take the assumed mean $\mathrm A = 64$ and $h = 1$, so $y_i = x_i – 64$:

The total $\sum f_iy_i = 0$, so the assumed mean was exactly right:

$$\bar x = 64 + \frac{0}{100} \times 1 = 64$$

$$\sigma^2 = \frac{1}{100^2}\left[100 \times 286 – 0^2\right] = \frac{28600}{10000} = 2.86$$

$$\sigma = \sqrt{2.86} \approx 1.69$$

Find the mean and variance for the following frequency distributions in questions 7 and 8.

Question 7

Solution. Midpoints: $15, 45, 75, 105, 135, 165, 195$; $\mathrm N = 30$.

$$\sum f_ix_i = 30 + 135 + 375 + 1050 + 405 + 825 + 390 = 3210$$

$$\bar x = \frac{3210}{30} = 107$$

$$\sum f_ix_i^2 = 450 + 6075 + 28125 + 110250 + 54675 + 136125 + 76050 = 411750$$

$$\sigma^2 = \frac{411750}{30} – 107^2 = 13725 – 11449 = 2276$$

Question 8

Solution. Midpoints: $5, 15, 25, 35, 45$; $\mathrm N = 50$.

$$\sum f_ix_i = 25 + 120 + 375 + 560 + 270 = 1350 \quad\Longrightarrow\quad \bar x = \frac{1350}{50} = 27$$

$$\sum f_ix_i^2 = 125 + 1800 + 9375 + 19600 + 12150 = 43050$$

$$\sigma^2 = \frac{43050}{50} – 27^2 = 861 – 729 = 132$$

Question 9

Find the mean, variance and standard deviation using the short-cut method.

Solution. Midpoints are $72.5, 77.5, \ldots, 112.5$. Take $\mathrm A = 92.5$ and $h = 5$, so $y_i = \dfrac{x_i – 92.5}{5}$:

$$\bar x = 92.5 + \frac{6}{60} \times 5 = 92.5 + 0.5 = 93$$

$$\sigma^2 = \frac{5^2}{60^2}\left[60 \times 254 – 6^2\right] = \frac{25}{3600}\left[15240 – 36\right] = \frac{25 \times 15204}{3600} = 105.583\ldots$$

$$\sigma = \sqrt{105.58} \approx 10.27$$

Question 10

The diameters of circles (in mm) drawn in a design are given below. Calculate the standard deviation and mean diameter of the circles.

Solution. The classes are discontinuous, so make them continuous first: $32.5$–$36.5$, $36.5$–$40.5$, $40.5$–$44.5$, $44.5$–$48.5$, $48.5$–$52.5$. Midpoints:

$$34.5,\ 38.5,\ 42.5,\ 46.5,\ 50.5$$

Take $\mathrm A = 42.5$ and $h = 4$, so $y_i = \dfrac{x_i – 42.5}{4}$:

$$\bar x = 42.5 + \frac{25}{100} \times 4 = 42.5 + 1 = 43.5$$

$$\sigma^2 = \frac{4^2}{100^2}\left[100 \times 199 – 25^2\right] = \frac{16}{10000}\left[19900 – 625\right] = \frac{16 \times 19275}{10000} = 30.84$$

$$\sigma = \sqrt{30.84} \approx 5.55$$

Common mistakes

• Confusing variance with standard deviation. $\sigma^2$ is the variance; $\sigma$ is its square root. Questions 1 to 5, 7 and 8 ask for the variance and questions 6, 9 and 10 for the standard deviation — read which is wanted.
• Question 2, quoting $\sum i^2$ wrongly. It is $\tfrac{n(n+1)(2n+1)}{6}$; using $\left(\tfrac{n(n+1)}{2}\right)^2$, which is $\sum i^3$, gives a quite different answer.
• Question 3, adding $3$ instead of multiplying. The multiples of $3$ are $3$ times the naturals, so the variance is multiplied by $3^2 = 9$. Adding a constant would have left the variance unchanged.
• Forgetting to subtract $\bar x^2$. The short formula is $\sigma^2 = \tfrac{\sum f_ix_i^2}{\mathrm N} – \bar x^2$; omitting the second term gives a number far too large.
• Questions 6, 9 and 10, dropping the $h^2$. The step-deviation variance must be multiplied by $h^2$ to return to the original units. With $h = 1$ in question 6 it makes no difference, which is exactly why it is easy to forget in questions 9 and 10.
• Question 10, using the classes as printed. $33$–$36$ and $37$–$40$ leave a gap, so the continuity correction is needed before the midpoints are read off.
• Rounding $\sqrt{\sigma^2}$ too early. In question 9, $\sigma^2 = 105.583\ldots$; rounding it to $105$ before the square root changes the answer’s second decimal.

Practise next

• Miscellaneous Exercise on Chapter 13 — recovering unknown observations from a given mean and variance, and correcting statistics after a data error.
• Exercise 13.1 — worth revisiting for the contrast: mean deviation uses moduli and cannot be manipulated algebraically, which is precisely why variance exists.