NCERT Class 11 Mathematics — Statistics, Miscellaneous Exercise on Chapter 13. All 6 questions solved.

Every question here runs the definitions backwards. Given

$$\bar x = \frac{\sum x_i}{n}, \qquad \sigma^2 = \frac{\sum x_i^2}{n} – \bar x^2$$

the two quantities that carry all the information are

$$\sum x_i = n\bar x, \qquad \sum x_i^2 = n\left(\sigma^2 + \bar x^2\right)$$

Once these two totals are known, an observation can be added, removed or replaced by adjusting them, and the new mean and variance read off.

Key insight. Work with the totals $\sum x_i$ and $\sum x_i^2$, never with the individual observations. Correcting a wrong reading is then just arithmetic on two numbers: subtract the wrong value (and its square) and add the right one. Questions 5 and 6 are that manoeuvre, and nothing else.

Question 1

The mean and variance of eight observations are $9$ and $9.25$ respectively. If six of the observations are $6, 7, 10, 12, 12$ and $13$, find the remaining two observations.

Solution. Let the missing observations be $a$ and $b$. From the two definitions:

$$\sum x_i = 8 \times 9 = 72$$

$$\sum x_i^2 = 8\left(9.25 + 81\right) = 8 \times 90.25 = 722$$

The six known values sum to

$$6 + 7 + 10 + 12 + 12 + 13 = 60$$

and their squares to

$$36 + 49 + 100 + 144 + 144 + 169 = 642$$

Therefore

$$a + b = 72 – 60 = 12 \qquad (1)$$ $$a^2 + b^2 = 722 – 642 = 80 \qquad (2)$$

From (1), $(a+b)^2 = 144$, so $a^2 + 2ab + b^2 = 144$. Subtracting (2):

$$2ab = 64 \quad\Longrightarrow\quad ab = 32$$

So $a$ and $b$ are the roots of $t^2 – 12t + 32 = 0$, that is $(t-4)(t-8) = 0$:

$$a = 4, \quad b = 8 \qquad\text{(or the other way round)}$$

Question 2

The mean and variance of $7$ observations are $8$ and $16$ respectively. If five of the observations are $2, 4, 10, 12, 14$, find the remaining two observations.

Solution. Same method.

$$\sum x_i = 7 \times 8 = 56, \qquad \sum x_i^2 = 7(16 + 64) = 560$$

The five known values sum to $42$ and their squares to

$$4 + 16 + 100 + 144 + 196 = 460$$

So

$$a + b = 14, \qquad a^2 + b^2 = 100$$

Then $(a+b)^2 = 196$ gives $2ab = 96$, so $ab = 48$, and $a$, $b$ are the roots of

$$t^2 – 14t + 48 = 0 \quad\Longrightarrow\quad (t-6)(t-8) = 0$$

Question 3

The mean and standard deviation of six observations are $8$ and $4$ respectively. If each observation is multiplied by $3$, find the new mean and new standard deviation of the resulting observations.

Solution. Multiplying every observation by a constant $k$ scales the mean by $k$ and the standard deviation by $|k|$ — the variance by $k^2$.

$$\text{new mean} = 3 \times 8 = 24$$

$$\text{new standard deviation} = 3 \times 4 = 12$$

(The number of observations plays no part.) This is the special case $a = 3$ of the general fact proved in question 4.

Question 4

Given that $\bar x$ is the mean and $\sigma^2$ is the variance of $n$ observations $x_1, x_2, \ldots, x_n$, prove that the mean and variance of the observations $ax_1, ax_2, \ldots, ax_n$ are $a\bar x$ and $a^2\sigma^2$ respectively, $a \ne 0$.

Solution. Write $y_i = ax_i$.

Mean: the constant factors straight out of the sum,

$$\bar y = \frac{1}{n}\sum_{i=1}^n ax_i = \frac an\sum_{i=1}^n x_i = a\bar x$$

Variance: substituting $\bar y = a\bar x$ into the definition,

$$\sigma_y^2 = \frac1n\sum_{i=1}^n\left(y_i – \bar y\right)^2 = \frac1n\sum_{i=1}^n\left(ax_i – a\bar x\right)^2$$

Factor $a$ out of the bracket and square it out of the sum:

$$= \frac1n\sum_{i=1}^n a^2\left(x_i – \bar x\right)^2 = a^2 \cdot \frac1n\sum_{i=1}^n\left(x_i – \bar x\right)^2 = a^2\sigma^2$$

$\blacksquare$

The squaring is what turns the factor $a$ into $a^2$ — and also why the standard deviation is multiplied by $|a|$ rather than $a$: it is a square root, so it cannot be negative.

Question 5

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) if the wrong item is omitted, (ii) if it is replaced by $12$.

Solution. First recover the two totals from the reported statistics:

$$\sum x_i = 20 \times 10 = 200$$

$$\sum x_i^2 = 20\left(2^2 + 10^2\right) = 20 \times 104 = 2080$$

(i) The wrong item omitted. Remove $8$ and $8^2 = 64$; the count drops to $19$:

$$\sum x_i = 200 – 8 = 192 \quad\Longrightarrow\quad \bar x = \frac{192}{19} = 10.105\ldots \approx 10.1$$

$$\sum x_i^2 = 2080 – 64 = 2016$$

$$\sigma^2 = \frac{2016}{19} – \left(\frac{192}{19}\right)^2 = 106.105\ldots – 102.116\ldots = 3.989\ldots$$

$$\sigma = \sqrt{3.989} \approx 1.997 \approx 1.99$$

(ii) Replaced by $12$. Remove $8$ and add $12$; the count stays at $20$:

$$\sum x_i = 200 – 8 + 12 = 204 \quad\Longrightarrow\quad \bar x = \frac{204}{20} = 10.2$$

$$\sum x_i^2 = 2080 – 64 + 144 = 2160$$

$$\sigma^2 = \frac{2160}{20} – (10.2)^2 = 108 – 104.04 = 3.96$$

$$\sigma = \sqrt{3.96} \approx 1.989 \approx 1.98$$

Question 6

The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3$ respectively. Later on it was found that three observations were incorrect, which were recorded as $21$, $21$ and $18$. Find the mean and standard deviation if the incorrect observations are omitted.

Solution. Recover the totals:

$$\sum x_i = 100 \times 20 = 2000$$

$$\sum x_i^2 = 100\left(3^2 + 20^2\right) = 100 \times 409 = 40900$$

The three incorrect readings sum to $21 + 21 + 18 = 60$, and their squares to

$$441 + 441 + 324 = 1206$$

Omitting them leaves $97$ observations:

$$\sum x_i = 2000 – 60 = 1940 \quad\Longrightarrow\quad \bar x = \frac{1940}{97} = 20$$

The mean is unchanged at exactly $20$ — the three discarded values happened to average $20$ themselves.

$$\sum x_i^2 = 40900 – 1206 = 39694$$

$$\sigma^2 = \frac{39694}{97} – 400 = 409.216\ldots – 400 = 9.216\ldots$$

$$\sigma = \sqrt{9.2165} \approx 3.036$$

The standard deviation rises slightly, because the discarded values were close to the mean and removing them leaves a relatively more spread-out set.

Common mistakes

• Questions 1 and 2, forgetting $\sum x_i^2 = n(\sigma^2 + \bar x^2)$. Using $n\sigma^2$ alone omits the $n\bar x^2$ term and makes the second equation unsolvable — or worse, solvable and wrong.
• Questions 1 and 2, solving $a + b$ and $a^2 + b^2$ by substitution. Squaring the sum and subtracting gives $2ab$ in one line, after which $a$ and $b$ are the roots of a quadratic.
• Question 3, scaling the standard deviation by $9$. The variance is multiplied by $3^2$; the standard deviation, being its square root, is multiplied by $3$.
• Question 3, confusing scaling with shifting. Adding $3$ to every observation would change the mean to $11$ and leave the standard deviation at $4$. Multiplying changes both.
• Question 5(i), keeping $n = 20$. Omitting an observation reduces the count to $19$, and that appears in both the mean and the variance.
• Question 5(ii), changing $n$. Replacing an observation leaves the count at $20$; only the two totals move.
• Subtracting the wrong value’s square as $8$ rather than $64$. The second total is of squares, so $8^2 = 64$ comes out and $12^2 = 144$ goes in.
• Question 6, assuming the mean must change. It does not here, because the three discarded values average exactly $20$. The standard deviation still changes, which is worth checking rather than assuming.

Practise next

• Exercise 14.1 — the next chapter, Probability, which begins a new topic.
• Exercise 13.2 — worth revisiting alongside questions 1, 2 and 5, since every one of them inverts the variance formula used there.