Statistics Practice Questions with Solutions, Class 10

Ques­tion 2#

The monthly elec­tric­ity use of 50 homes is given. Find the mean by the assumed mean method.

Solu­tion. Take $a = 100$, the cen­tral class mark. The last two columns are filled in the table above.

$$\displaystyle \bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i} = 100 + \dfrac{140}{50} = 100 + 2.8 = 102.8$$

Answer: the mean use is 102.8 units.

Ques­tion 3#

The monthly spend­ing on snacks by 50 stu­dents is given. Find the mean by the step-devi­a­tion method.

Solu­tion. Class marks: 175, 225, 275, 325, 375, 425. Take $a = 275$ and $h = 50$, so $u_i = -2, -1, 0, 1, 2, 3$.

$$\displaystyle \begin{aligned} \sum f_i u_i &= 4(-2) + 9(-1) + 15(0) + 11(1) + 7(2) + 4(3) \\ &= -8 - 9 + 0 + 11 + 14 + 12 = 20 \\ \bar{x} &= 275 + 50 \times \dfrac{20}{50} = 275 + 20 = 295 \end{aligned}$$

Answer: the mean spend­ing is ₹295.

Why this works: all the classes are 50 wide, so divid­ing every devi­a­tion by 50 gives whole num­bers and does not change the mean once we mul­ti­ply back by $h$.

Ques­tion 4#

The mean of the dis­tri­b­u­tion below is 23.5. Find $p$.

Solu­tion. Class marks: 5, 15, 25, 35, 45.

$$\displaystyle \begin{aligned} \sum f_i &= 32 + p \\ \sum f_i x_i &= 35 + 165 + 25p + 315 + 225 = 740 + 25p \\ \dfrac{740 + 25p}{32 + p} &= 23.5 \\ 740 + 25p &= 752 + 23.5p \\ 1.5p &= 12 \\ p &= 8 \end{aligned}$$

Check: $\displaystyle \frac{740 + 200}{40} = \frac{940}{40} = 23.5$.

Answer: $p = 8$.

Ques­tion 5#

The dis­tri­b­u­tion below has 30 obser­va­tions and a mean of 34. Find $a$ and $b$.

Solu­tion. From the total, $4 + a + 10 + b + 3 = 30$, so $a + b = 13$.

With class marks 15, 25, 35, 45, 55:

$$\begin{aligned} 60 + 25a + 350 + 45b + 165 &= 34 \times 30 \\ 575 + 25a + 45b &= 1020 \\ 25a + 45b &= 445 \\ 5a + 9b &= 89 \end{aligned}$$

Sub­sti­tut­ing $a = 13 - b$: $65 - 5b + 9b = 89$, so $4b = 24$ and $b = 6$. Then $a = 7$.

Check: $60 + 175 + 350 + 270 + 165 = 1020$ and $\displaystyle \frac{1020}{30} = 34$.

Answer: $a = 7$, $b = 6$.

Mode of grouped data#

Ques­tion 6#

The daily dis­tance cycled by 70 rid­ers is shown. Find the modal dis­tance.

Solu­tion. The high­est fre­quency is 22, so the modal class is 40–60: $l = 40$, $h = 20$, $f_1 = 22$, $f_0 = 14$, $f_2 = 16$.

$$\displaystyle \begin{aligned} \text{Mode} &= 40 + \dfrac{22 - 14}{44 - 14 - 16} \times 20 \\ &= 40 + \dfrac{8}{14} \times 20 = 40 + \dfrac{160}{14} \approx 40 + 11.43 = 51.43 \end{aligned}$$

Answer: the modal dis­tance is about 51.43 km.

Ques­tion 7#

Find the mode of the dis­tri­b­u­tion below.

Solu­tion. The modal class is the last class, 30–40. There is no class after it, so $f_2 = 0$. Also $l = 30$, $h = 10$, $f_1 = 11$, $f_0 = 8$.

$$\displaystyle \text{Mode} = 30 + \dfrac{11 - 8}{22 - 8 - 0} \times 10 = 30 + \dfrac{30}{14} \approx 30 + 2.14 = 32.14$$

Answer: the mode is about 32.14.

Why this works: a miss­ing neigh­bour con­tributes a fre­quency of zero.

Median of grouped data#

Ques­tion 8#

The weights of 50 peo­ple are shown. Find the median weight.

Solu­tion. The cumu­la­tive fre­quen­cies are 5, 13, 27, 39, 46, 50. Here $n = 50$ and $\displaystyle \frac{n}{2} = 25$. The first $cf$ reach­ing 25 is 27, so the median class is 40–50, with $l = 40$, $cf = 13$, $f = 14$, $h = 10$.

$$\displaystyle \text{Median} = 40 + \dfrac{25 - 13}{14} \times 10 = 40 + \dfrac{120}{14} \approx 40 + 8.57 = 48.57$$

Answer: the median weight is about 48.57 kg.

Ques­tion 9#

The scores of 50 play­ers in a game are given. Find the median score.

Solu­tion. The classes are inclu­sive. Half the gap is $\displaystyle \frac{40 - 39}{2} = 0.5$, so the classes become 29.5–39.5, 39.5–49.5, 49.5–59.5, 59.5–69.5, 69.5–79.5.

The cumu­la­tive fre­quen­cies are 4, 15, 31, 44, 50. $\displaystyle \frac{n}{2} = 25$, so the median class is 49.5–59.5, with $l = 49.5$, $cf = 15$, $f = 16$, $h = 10$.

$$\displaystyle \text{Median} = 49.5 + \dfrac{25 - 15}{16} \times 10 = 49.5 + 6.25 = 55.75$$

Answer: the median score is 55.75.

Ques­tion 10#

The median of the dis­tri­b­u­tion below is 22.5 and the total fre­quency is 40. Find $x$ and $y$.

Solu­tion. The median 22.5 lies in 20–30, so this is the median class: $l = 20$, $f = 12$, $h = 10$, $\displaystyle \frac{n}{2} = 20$, and $cf = 6 + x$.

$$\displaystyle \begin{aligned} 20 + \dfrac{20 - (6 + x)}{12} \times 10 &= 22.5 \\ \dfrac{(14 - x) \times 10}{12} &= 2.5 \\ 14 - x &= 3 \\ x &= 11 \end{aligned}$$

From the total: $6 + 11 + 12 + y + 5 = 40$, so $y = 6$.

Check: the cumu­la­tive fre­quen­cies are 6, 17, 29, 35, 40; the first to reach 20 is 29, in 20–30, and $\displaystyle 20 + \frac{3}{12} \times 10 = 22.5$.

Answer: $x = 11$, $y = 6$.

Empir­i­cal rela­tion and exam-level ques­tions#

Ques­tion 11#

For a dis­tri­b­u­tion, the mode is 44 and the mean is 47. Use the empir­i­cal rela­tion­ship to find the median.

Solu­tion.

$$\displaystyle \begin{aligned} 3\,\text{Median} &= \text{Mode} + 2\,\text{Mean} = 44 + 2(47) = 138 \\ \text{Median} &= \dfrac{138}{3} = 46 \end{aligned}$$

Answer: the median is about 46.

Why this works: the median of a mod­er­ately lop­sided dis­tri­b­u­tion usu­ally lies between the mode and the mean, about two-thirds of the way from the mode to the mean.

Ques­tion 12#

The marks of 50 stu­dents are given. Find the mean, median and mode, and check the empir­i­cal rela­tion­ship.

Solu­tion. The work­ing columns are included in the table.

Mean:

$$\displaystyle \bar{x} = \dfrac{1500}{50} = 30$$

Median: $\displaystyle \frac{n}{2} = 25$; the first $cf$ reach­ing 25 is 39, so the median class is 30–40 with $l = 30$, $cf = 24$, $f = 15$, $h = 10$.

$$\displaystyle \text{Median} = 30 + \dfrac{25 - 24}{15} \times 10 = 30 + \dfrac{2}{3} \approx 30.67$$

Mode: the modal class is also 30–40, with $f_1 = 15$, $f_0 = 12$, $f_2 = 7$.

$$\displaystyle \text{Mode} = 30 + \dfrac{15 - 12}{30 - 12 - 7} \times 10 = 30 + \dfrac{30}{11} \approx 32.73$$

Check:

$$\displaystyle \begin{aligned} 3\,\text{Median} &= 3 \times \dfrac{92}{3} = 92 \\ \text{Mode} + 2\,\text{Mean} &\approx 32.73 + 60 = 92.73 \end{aligned}$$

The two sides agree to within 1, so the rela­tion holds approx­i­mately. It would esti­mate the mode as $92 - 60 = 32$, close to the for­mula value 32.73.

Answer: mean = 30, median ≈ 30.67, mode ≈ 32.73; the empir­i­cal rela­tion holds approx­i­mately.

Key terms#

Class mark

The mid-point of a class inter­val.

Assumed mean

A con­ve­nient cen­tral value from which devi­a­tions are mea­sured.

Step devi­a­tion

A devi­a­tion divided by the class size, $\displaystyle u_i = \frac{x_i - a}{h}$.

Modal class

The class with the high­est fre­quency.

Cumu­la­tive fre­quency

The run­ning total of fre­quen­cies up to a class.

Median class

The first class whose cumu­la­tive fre­quency is at least $\displaystyle \frac{n}{2}$.

Class bound­aries

The lim­its of con­tin­u­ous classes made from inclu­sive classes by mov­ing each limit half the gap.

Empir­i­cal rela­tion­ship

The approx­i­mate rule $3\,\text{Median} = \text{Mode} + 2\,\text{Mean}$.

Com­mon ques­tions#

Which mean method should I use if the ques­tion does not say?#

Any method earns full marks. Use step devi­a­tion for large class marks with equal widths, and the direct method when there is a miss­ing fre­quency.

Are the modal class and the median class always the same?#

No. They often coin­cide, as in Ques­tion 12, but the modal class depends on the largest fre­quency while the median class depends on $\displaystyle \frac{n}{2}$ and the cumu­la­tive fre­quen­cies.

How do I check a miss­ing fre­quency answer?#

Put the value back into the table, recom­pute the mean or median, and con­firm that the class you assumed is still the median class or modal class.

Do I round in the mid­dle of the work­ing?#

Keep frac­tions such as $\displaystyle \frac{160}{14}$ until the last step, then round to two dec­i­mal places. Early round­ing can change the final digit.

Is the ogive part of this chap­ter?#

No. The ratio­nalised Class 10 syl­labus finds the median from the for­mula, not from a cumu­la­tive fre­quency graph.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.