NCERT Exemplar Solutions Class 11 Maths Statistics — Objective Type and Fill in the Blanks Q24–46

NCERT Exemplar Problems, Class 11 Mathematics — Statistics, Objective Type and Fill in the Blanks questions 24 to 46. All 23 questions solved.

Sixteen multiple-choice questions and seven fill-in-the-blanks. About half are direct calculations (Q24, Q26 to Q28, Q30, Q34, Q37) and the rest test how mean deviation and standard deviation behave when the data are shifted, scaled or transformed (Q31 to Q33, Q35, Q36, Q38, Q39, Q41 to Q46). Those need only two facts: adding a constant leaves the spread unchanged, and multiplying by $k$ multiplies the standard deviation by $|k|$.

One answer in the printed key is wrong, Question 26. It is explained where it occurs.

Key insight. Spread is blind to where the data sit but not to how stretched they are. Shifting $x\to x+k$ changes the mean but not $\sigma$; scaling $x\to lx$ multiplies $\sigma$ by $|l|$ and the variance by $l^2$. Most of the second half of this page is that one rule.

Question 24

The mean deviation of the data $3, 10, 10, 4, 7, 10, 5$ from the mean is

Solution. The sum is $49$ over $7$ values, so the mean is $7$. The deviations are $4,3,3,3,0,3,2$, with sum $18$. The mean deviation is $\dfrac{18}{7}\approx2.57$.

Question 25

Mean deviation for $n$ observations $x_1,x_2,\dots,x_n$ from their mean $\bar x$ is given by

Solution. The mean deviation is the average of the absolute deviations. Option (A) is always $0$, since deviations from the mean cancel; (D) is the variance; (C) is $n$ times the variance.

Question 26

When tested, the lives (in hours) of $5$ bulbs were noted as follows: $1357, 1090, 1666, 1494, 1623$. The mean deviations (in hours) from their mean is

Solution. The sum is $1357+1090+1666+1494+1623=7230$, so the mean is $1446$. The absolute deviations are $|1357-1446|=89$, $|1090-1446|=356$, $|1666-1446|=220$, $|1494-1446|=48$, $|1623-1446|=177$. Their sum is $89+356+220+48+177=890$, so the mean deviation is $\dfrac{890}{5}=178$.

The printed key says (B) $179$; that is wrong. The sum $890$ is exactly divisible by $5$, giving exactly $178$, and nothing in the data produces $179$.

Question 27

Following are the marks obtained by $9$ students in a mathematics test: $50, 69, 20, 33, 53, 39, 40, 65, 59$. The mean deviation from the median is:

Solution. Arrange in order: $20,33,39,40,50,53,59,65,69$. The median is the 5th value, $50$. The deviations are $30,17,11,10,0,3,9,15,19$, sum $114$, and $\dfrac{114}{9}=12.67$.

Question 28

The standard deviation of the data $6,5,9,13,12,8,10$ is

Solution. The sum is $63$ over $7$ values, so the mean is $9$. The squared deviations are $9,16,0,16,9,1,1$, adding to $52$. So $\sigma^2=\dfrac{52}{7}$ and $\sigma=\sqrt{\dfrac{52}{7}}$. Option (B) is the variance, not the standard deviation.

Question 29

Let $x_1,x_2,\dots,x_n$ be $n$ observations and $\bar x$ be their arithmetic mean. The formula for the standard deviation is given by

Solution. The standard deviation is the square root of the variance, $\dfrac{\sum(x_i-\bar x)^2}{n}$. Option (B) is the variance itself. Option (D) has a plus sign, but the shortcut formula subtracts: $\sqrt{\dfrac{\sum x_i^2}{n}-\bar x^2}$.

Question 30

The mean of $100$ observations is $50$ and their standard deviation is $5$. The sum of all squares of all the observations is

Solution. $\sum x^2=N(\sigma^2+\bar x^2)=100(25+2500)=252500$.

Question 31

Let $a,b,c,d,e$ be the observations with mean $m$ and standard deviation $s$. The standard deviation of the observations $a+k,b+k,c+k,d+k,e+k$ is

Solution. The mean becomes $m+k$, so every deviation $(x+k)-(m+k)=x-m$ is unchanged, and so is the standard deviation.

Question 32

Let $x_1,x_2,x_3,x_4,x_5$ be the observations with mean $m$ and standard deviation $s$. The standard deviation of the observations $kx_1,kx_2,kx_3,kx_4,kx_5$ is

Solution. The mean becomes $km$ and each deviation becomes $k(x-m)$, so the variance is multiplied by $k^2$ and the standard deviation by $|k|$. For the positive $k$ the question has in mind this is $ks$.

Question 33

Let $x_1,x_2,\dots,x_n$ be $n$ observations. Let $w_i=lx_i+k$ for $i=1,2,\dots,n$, where $l$ and $k$ are constants. If the mean of the $x_i$’s is $48$ and their standard deviation is $12$, the mean of the $w_i$’s is $55$ and the standard deviation of the $w_i$’s is $15$, the values of $l$ and $k$ should be

Solution. The standard deviation scales by $|l|$, so $|l|\times12=15$ and $|l|=1.25$. The mean transforms the same way as each value, so $48l+k=55$. With $l=1.25$: $60+k=55$, giving $k=-5$. With $l=-1.25$ we would need $k=115$, which is not offered. Options (C) and (D) fail the standard-deviation condition, and (B) gives mean $-55$.

Question 34

Standard deviations for first $10$ natural numbers is

Solution. From the unit’s own result, $\sigma=\sqrt{\dfrac{n^2-1}{12}}=\sqrt{\dfrac{99}{12}}=\sqrt{8.25}\approx2.87$.

Question 35

Consider the numbers $1,2,3,4,5,6,7,8,9,10$. If $1$ is added to each number, the variance of the numbers so obtained is

Solution. Adding a constant does not change the variance, so it stays the variance of $1,\dots,10$, which is $\dfrac{10^2-1}{12}=8.25$. The $2.87$ in option (B) is the standard deviation, a common trap.

Question 36

Consider the first $10$ positive integers. If we multiply each number by $-1$ and then add $1$ to each number, the variance of the numbers so obtained is

Solution. The transformation is $x\to1-x$. The shift by $1$ changes nothing, and multiplying by $-1$ multiplies the variance by $(-1)^2=1$. So the variance remains $8.25$.

Question 37

The following information relates to a sample of size $60$: $\sum x^2=18000$, $\sum x=960$. The variance is

Solution. $\bar x=\dfrac{960}{60}=16$ and $\dfrac{\sum x^2}{N}=\dfrac{18000}{60}=300$, so $\sigma^2=300-16^2=300-256=44$. (Option (A) is $\sqrt{44}$, the standard deviation.)

Question 38

Coefficient of variation of two distributions are $50$ and $60$, and their arithmetic means are $30$ and $25$ respectively. Difference of their standard deviation is

Solution. From $\text{C.V.}=\dfrac{\sigma}{\bar x}\times100$ we get $\sigma=\dfrac{\text{C.V.}\times\bar x}{100}$. So $\sigma_1=\dfrac{50\times30}{100}=15$ and $\sigma_2=\dfrac{60\times25}{100}=15$. The difference is $0$. (Coefficient of variation is no longer in the current NCERT Class 11 syllabus, but the question still appears in this book.)

Question 39

The standard deviation of some temperature data in $^\circ$C is $5$. If the data were converted into $^\circ$F, the variance would be

Solution. The conversion is $F=\dfrac95C+32$. The $+32$ does not affect spread, and the factor $\dfrac95$ multiplies the standard deviation, giving $\dfrac95\times5=9$. The variance is $9^2=81$.

Fill in the blanks in Questions 40 to 46.

Question 40

Coefficient of variation $=\dfrac{\_\_\_\_}{\text{Mean}}\times100$

Solution. The coefficient of variation compares spread with the mean, so the numerator is the standard deviation. (This measure is not in the current Class 11 syllabus.)

Question 41

If $\bar x$ is the mean of $n$ values of $x$, then $\sum_{i=1}^n(x_i-\bar x)$ is always equal to ______. If $a$ has any value other than $\bar x$, then $\sum_{i=1}^n(x_i-\bar x)^2$ is ______ than $\sum(x_i-a)^2$.

Solution. The first sum is $\sum x_i-n\bar x=0$. For the second, $\sum(x_i-a)^2=\sum(x_i-\bar x)^2+n(\bar x-a)^2$, and the extra term is positive whenever $a\ne\bar x$. So the sum of squares about the mean is smaller: the mean is the point that minimises it.

Question 42

If the variance of a data is $121$, then the standard deviation of the data is ______.

Solution. $\sigma=\sqrt{121}=11$.

Question 43

The standard deviation of a data is ______ of any change in origin, but is ______ on the change of scale.

Solution. Shifting the origin ($x\to x+k$) leaves $\sigma$ unchanged, while rescaling ($x\to lx$) multiplies it by $|l|$. See Questions 31 and 32.

Question 44

The sum of the squares of the deviations of the values of the variable is ______ when taken about their arithmetic mean.

Solution. By Question 41, $\sum(x_i-a)^2$ is smallest when $a=\bar x$.

Question 45

The mean deviation of the data is ______ when measured from the median.

Solution. $\sum|x_i-a|$ is smallest when $a$ is the median (moving $a$ across the median makes more points farther than nearer), so the mean deviation about the median is the least.

Question 46

The standard deviation is ______ to the mean deviation taken from the arithmetic mean.

Solution. By the Cauchy–Schwarz inequality, $\left(\dfrac1n\sum|d_i|\right)^2\le\dfrac1n\sum d_i^2$, so $\sigma\ge$ mean deviation about the mean, with equality only when all the $|d_i|$ are equal (for example the data $1,3$). Check with Question 1 of the previous page: $\sigma\approx1.46$ against a mean deviation of $1.25$.

Common mistakes

• Question 26: a slip in one absolute value. Five deviations of different size are easy to mis-subtract; adding the signed deviations (they should cancel to $0$) catches it before the sum is divided.
• Question 27: finding the median before ordering. With nine values the median is the fifth after sorting, not the fifth as listed.
• Questions 28, 35 and 37: reporting variance for standard deviation or vice versa. The distractor options are exactly this: $\dfrac{52}{7}$, $2.87$ and $6.63$ are the “other” quantity. Check which one the question asks for.
• Question 31 and 33: adding $k$ to the standard deviation. A shift moves the mean, never the spread; only the multiplier changes $\sigma$, and it enters as $|l|$.
• Question 36: giving a negative variance, or $-8.25$. Multiplying by $-1$ squares to $+1$ in the variance, so the answer is unchanged.
• Question 39: squaring before scaling the wrong quantity. The standard deviation is multiplied by $\dfrac95$, so the variance is multiplied by $\dfrac{81}{25}$; students multiply the variance by $\dfrac95$ only.
• Question 41: the second blank. Many answer “greater than” by assuming a different reference point must give a bigger sum; the mean is the minimiser, so the sum about the mean is smaller.

Practise next

• The calculation questions behind these answers are on Short and Long Answer Type, Q1–23.
• NCERT Class 11 Chapter 13: Exercise 13.1, Exercise 13.2 and the Miscellaneous Exercise.