Textbook - Statistics Questions in English

(A) To find the mean marks,we calculate the product of each mark $(x_{i})$ with its corresponding frequency $(f_{i})$.

Marks obtained $(x_{i})$	Number of students $(f_{i})$	$f_{i}x_{i}$
$10$	$1$	$10$
$20$	$1$	$20$
$36$	$3$	$108$
$40$	$4$	$160$
$50$	$3$	$150$
$56$	$2$	$112$
$60$	$4$	$240$
$70$	$4$	$280$
$72$	$1$	$72$
$80$	$1$	$80$
$88$	$2$	$176$
$92$	$3$	$276$
$95$	$1$	$95$
Total	$\Sigma f_{i} = 30$	$\Sigma f_{i}x_{i} = 1779$

The mean $\bar{x}$ is given by the formula:
$\bar{x} = \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}} = \frac{1779}{30} = 59.3$
Therefore,the mean marks obtained by the students is $59.3$.

(B) To find the mean,we first determine the class marks $(x_i)$ for each interval.
Let the assumed mean $a = 50$ and class size $h = 10$. Then $d_i = x_i - 50$ and $u_i = \frac{x_i - 50}{10}$.

Percentage of female teachers	Number of states/$U$.$T$. $(f_i)$	$x_i$	$d_i = x_i - 50$	$u_i = \frac{x_i - 50}{10}$	$f_i x_i$	$f_i d_i$	$f_i u_i$
$15$-$25$	$6$	$20$	-$30$	-$3$	$120$	-$180$	-$18$
$25$-$35$	$11$	$30$	-$20$	-$2$	$330$	-$220$	-$22$
$35$-$45$	$7$	$40$	-$10$	-$1$	$280$	-$70$	-$7$
$45$-$55$	$4$	$50$	$0$	$0$	$200$	$0$	$0$
$55$-$65$	$4$	$60$	$10$	$1$	$240$	$40$	$4$
$65$-$75$	$2$	$70$	$20$	$2$	$140$	$40$	$4$
$75$-$85$	$1$	$80$	$30$	$3$	$80$	$30$	$3$
Total	$35$	-	-	-	$1390$	-$360$	-$36$

From the table,we have $\Sigma f_i = 35, \Sigma f_i x_i = 1390, \Sigma f_i d_i = -360, \Sigma f_i u_i = -36$.
$1$. Direct Method: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1390}{35} \approx 39.71$.
$2$. Assumed Mean Method: $\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 50 + \frac{-360}{35} = 50 - 10.29 = 39.71$.
$3$. Step-deviation Method: $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h = 50 + \left(\frac{-36}{35}\right) \times 10 = 50 - 10.29 = 39.71$.
The mean percentage of female teachers is $39.71$.

(C) To find the mean,we first calculate the class marks $(x_i)$ for each interval.

Number of wickets	Number of bowlers $(f_i)$	Class mark $(x_i)$	$f_i x_i$
$20$-$60$	$7$	$40$	$280$
$60$-$100$	$5$	$80$	$400$
$100$-$150$	$16$	$125$	$2000$
$150$-$250$	$12$	$200$	$2400$
$250$-$350$	$2$	$300$	$600$
$350$-$450$	$3$	$400$	$1200$
Total	$\sum f_i = 45$		$\sum f_i x_i = 6880$

The mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{6880}{45} \approx 152.89$.
This mean signifies that,on average,the bowlers have taken $152.89$ wickets in the given one-day cricket matches.

(D) To find the class mark $(x_i)$ for each interval,the following relation is used:
Class mark $(x_i) = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
The values of $x_i$ and $f_ix_i$ are calculated as follows:

Number of plants	Number of houses $(f_i)$	$x_i$	$f_ix_i$
$0-2$	$1$	$1$	$1 \times 1 = 1$
$2-4$	$2$	$3$	$2 \times 3 = 6$
$4-6$	$1$	$5$	$1 \times 5 = 5$
$6-8$	$5$	$7$	$5 \times 7 = 35$
$8-10$	$6$	$9$	$6 \times 9 = 54$
$10-12$	$2$	$11$	$2 \times 11 = 22$
$12-14$	$3$	$13$	$3 \times 13 = 39$
Total	$\sum f_i = 20$		$\sum f_ix_i = 162$

From the table,we have $\sum f_i = 20$ and $\sum f_ix_i = 162$.
The mean $\bar{x}$ is given by $\bar{x} = \frac{\sum f_ix_i}{\sum f_i}$.
Mean $= \frac{162}{20} = 8.1$.
Therefore,the mean number of plants per house is $8.1$.
The direct method is used here because the numerical values of $x_i$ and $f_i$ are small,making the calculations simple.

(A) To find the class mark $(x_i)$ for each interval,the following relation is used:
$x_i = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
Class size $(h)$ of this data $= 20$.
Taking $150$ as the assumed mean $(a)$,$d_i$,$u_i$,and $f_i u_i$ can be calculated as follows:
$\begin{array}{|c|c|c|c|c|c|} \hline \text{Daily wages} & f_i & x_i & d_i = x_i - 150 & u_i = \frac{d_i}{20} & f_i u_i \\ \hline 100-120 & 12 & 110 & -40 & -2 & -24 \\ \hline 120-140 & 14 & 130 & -20 & -1 & -14 \\ \hline 140-160 & 8 & 150 & 0 & 0 & 0 \\ \hline 160-180 & 6 & 170 & 20 & 1 & 6 \\ \hline 180-200 & 10 & 190 & 40 & 2 & 20 \\ \hline \text{Total} & 50 & & & & -12 \\ \hline \end{array}$
From the table,we have:
$\sum f_i = 50$
$\sum f_i u_i = -12$
Using the step-deviation method,the mean $\bar{x}$ is given by:
$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$
$\bar{x} = 150 + \left( \frac{-12}{50} \right) \times 20$
$\bar{x} = 150 - \frac{240}{50} = 150 - 4.8$
$\bar{x} = 145.2$
Therefore,the mean daily wage of the workers is $Rs. 145.20$.

(B) To find the class mark $(x_{i})$ for each interval,the following relation is used:
$x_{i} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
Given that,mean pocket allowance,$\bar{x} = Rs. 18$.
Taking $18$ as assumed mean $(a)$,$d_{i}$ and $f_{i}d_{i}$ are calculated as follows:

Daily pocket allowance (in $Rs.$)	Number of children $(f_{i})$	Class mark $(x_{i})$	$d_{i} = x_{i} - 18$	$f_{i}d_{i}$
$11-13$	$7$	$12$	$-6$	$-42$
$13-15$	$6$	$14$	$-4$	$-24$
$15-17$	$9$	$16$	$-2$	$-18$
$17-19$	$13$	$18$	$0$	$0$
$19-21$	$f$	$20$	$2$	$2f$
$21-23$	$5$	$22$	$4$	$20$
$23-25$	$4$	$24$	$6$	$24$
Total	$\sum f_{i} = 44 + f$			$\sum f_{i}d_{i} = 2f - 40$

From the table,we obtain:
$\sum f_{i} = 44 + f$
$\sum f_{i}d_{i} = 2f - 40$
Using the formula for mean:
$\bar{x} = a + \frac{\sum f_{i}d_{i}}{\sum f_{i}}$
$18 = 18 + \left(\frac{2f - 40}{44 + f}\right)$
$0 = \frac{2f - 40}{44 + f}$
$2f - 40 = 0$
$2f = 40$
$f = 20$
Hence,the missing frequency $f$ is $20$.

(C) To find the class mark $(x_i)$ of each interval,the following formula is used:
$x_i = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
The class size $(h)$ of this data is $3$.
Taking $75.5$ as the assumed mean $(a)$,we calculate $d_i, u_i,$ and $f_i u_i$ as follows:

Number of heartbeats per minute	Number of women $(f_i)$	$x_i$	$d_i = x_i - 75.5$	$u_i = \frac{d_i}{3}$	$f_i u_i$
$65$-$68$	$2$	$66.5$	-$9$	-$3$	-$6$
$68$-$71$	$4$	$69.5$	-$6$	-$2$	-$8$
$71$-$74$	$3$	$72.5$	-$3$	-$1$	-$3$
$74$-$77$	$8$	$75.5$	$0$	$0$	$0$
$77$-$80$	$7$	$78.5$	$3$	$1$	$7$
$80$-$83$	$4$	$81.5$	$6$	$2$	$8$
$83$-$86$	$2$	$84.5$	$9$	$3$	$6$
Total	$30$	-	-	-	$4$

From the table,we have $\sum f_i = 30$ and $\sum f_i u_i = 4$.
The mean $\bar{x}$ is calculated using the step-deviation method:
$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$
$\bar{x} = 75.5 + \left( \frac{4}{30} \right) \times 3$
$\bar{x} = 75.5 + 0.4 = 75.9$
Therefore,the mean heartbeats per minute for these women is $75.9$.

(D) The class intervals are not continuous. The gap between two consecutive class intervals is $1$. To make them continuous,we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of each class.

Class interval	$f_i$	$x_i$	$d_i = x_i - 57$	$u_i = d_i / 3$	$f_i u_i$
$49.5-52.5$	$15$	$51$	$-6$	$-2$	$-30$
$52.5-55.5$	$110$	$54$	$-3$	$-1$	$-110$
$55.5-58.5$	$135$	$57$	$0$	$0$	$0$
$58.5-61.5$	$115$	$60$	$3$	$1$	$115$
$61.5-64.5$	$25$	$63$	$6$	$2$	$50$
Total	$400$				$25$

Here,$\sum f_i = 400$,$\sum f_i u_i = 25$,assumed mean $a = 57$,and class size $h = 3$.
Using the step-deviation method:
$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$
$\bar{x} = 57 + \left( \frac{25}{400} \right) \times 3 = 57 + \frac{75}{400} = 57 + 0.1875 = 57.1875 \approx 57.19$.
We chose the step-deviation method because the values of $f_i$ and $d_i$ are large and there is a common factor $h=3$.

(A) To find the class mark $(x_{i})$ for each interval,the following relation is used:
$x_{i} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
Class size $(h) = 50$.
Taking $225$ as the assumed mean $(a)$,we calculate $d_i$,$u_i$,and $f_iu_i$ as follows:

Daily expenditure (in $Rs.$)	$f_{i}$	$x_{i}$	$d_{i} = x_{i} - 225$	$u_{i} = \frac{d_{i}}{50}$	$f_{i}u_{i}$
$100-150$	$4$	$125$	$-100$	$-2$	$-8$
$150-200$	$5$	$175$	$-50$	$-1$	$-5$
$200-250$	$12$	$225$	$0$	$0$	$0$
$250-300$	$2$	$275$	$50$	$1$	$2$
$300-350$	$2$	$325$	$100$	$2$	$4$
Total	$\Sigma f_{i} = 25$				$\Sigma f_{i}u_{i} = -7$

Using the step-deviation method:
Mean $(\bar{x}) = a + \left(\frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}\right) \times h$
$\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50$
$\bar{x} = 225 + (-7 \times 2)$
$\bar{x} = 225 - 14 = 211$
Therefore,the mean daily expenditure on food is $Rs. 211$.

(B) To find the class marks $(x_{i})$ for each interval,we use the formula:
$x_{i} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
The class size $(h)$ is $0.04 - 0.00 = 0.04$.
Let the assumed mean $(a)$ be $0.14$.
We calculate $d_{i} = x_{i} - a$ and $u_{i} = \frac{d_{i}}{h}$.

Concentration of $SO_{2}$ $(ppm)$	Frequency $(f_{i})$	Class mark $(x_{i})$	$d_{i} = x_{i} - 0.14$	$u_{i} = \frac{d_{i}}{0.04}$	$f_{i}u_{i}$
$0.00-0.04$	$4$	$0.02$	$-0.12$	$-3$	$-12$
$0.04-0.08$	$9$	$0.06$	$-0.08$	$-2$	$-18$
$0.08-0.12$	$9$	$0.10$	$-0.04$	$-1$	$-9$
$0.12-0.16$	$2$	$0.14$	$0$	$0$	$0$
$0.16-0.20$	$4$	$0.18$	$0.04$	$1$	$4$
$0.20-0.24$	$2$	$0.22$	$0.08$	$2$	$4$
Total	$\Sigma f_{i} = 30$				$\Sigma f_{i}u_{i} = -31$

Using the step-deviation method:
Mean $\bar{x} = a + \left( \frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}} \right) \times h$
$\bar{x} = 0.14 + \left( \frac{-31}{30} \right) \times 0.04$
$\bar{x} = 0.14 - \frac{1.24}{30}$
$\bar{x} = 0.14 - 0.04133...$
$\bar{x} \approx 0.09867 \approx 0.099\, ppm$.
Thus,the mean concentration of $SO_{2}$ in the air is $0.099\, ppm$.

(C) To find the mean number of days,we first calculate the class mark $(x_i)$ for each interval using the formula: $x_i = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$.
Using the assumed mean method with $a = 17$,we calculate $d_i = x_i - a$ and $f_i d_i$:

Number of days	Number of students $(f_i)$	$x_i$	$d_i = x_i - 17$	$f_i d_i$
$0-6$	$11$	$3$	$-14$	$-154$
$6-10$	$10$	$8$	$-9$	$-90$
$10-14$	$7$	$12$	$-5$	$-35$
$14-20$	$4$	$17$	$0$	$0$
$20-28$	$4$	$24$	$7$	$28$
$28-38$	$3$	$33$	$16$	$48$
$38-40$	$1$	$39$	$22$	$22$
Total	$\sum f_i = 40$			$\sum f_i d_i = -181$

The mean $\bar{x}$ is given by $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$.
Substituting the values: $\bar{x} = 17 + \frac{-181}{40} = 17 - 4.525 = 12.475$.
Rounding to two decimal places,the mean number of days is $12.48$ days.

(D) To find the class marks $(x_i)$,the following relation is used:
$x_i = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
Class size $(h)$ for this data $= 10$.
Taking $70$ as the assumed mean $(a)$,$d_i$,$u_i$,and $f_i u_i$ are calculated as follows:

Literacy rate (in $\%$)	Number of cities $(f_i)$	$x_i$	$d_i = x_i - 70$	$u_i = \frac{d_i}{10}$	$f_i u_i$
$45-55$	$3$	$50$	$-20$	$-2$	$-6$
$55-65$	$10$	$60$	$-10$	$-1$	$-10$
$65-75$	$11$	$70$	$0$	$0$	$0$
$75-85$	$8$	$80$	$10$	$1$	$8$
$85-95$	$3$	$90$	$20$	$2$	$6$
Total	$\sum f_i = 35$				$\sum f_i u_i = -2$

From the table,we obtain:
$\sum f_i = 35$
$\sum f_i u_i = -2$
Mean,$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$
$= 70 + \left(\frac{-2}{35}\right) \times 10$
$= 70 - \frac{20}{35}$
$= 70 - \frac{4}{7}$
$= 70 - 0.5714...$
$\approx 69.43$
Therefore,the mean literacy rate is $69.43 \%$.

(A) To find the mode,we first organize the data into a frequency distribution table:

Number of wickets	$0$	$1$	$2$	$3$	$4$	$5$	$6$
Number of matches	$1$	$1$	$3$	$2$	$1$	$1$	$1$

The mode is the observation that occurs most frequently. From the table,the number of wickets $2$ appears $3$ times,which is the highest frequency.
Therefore,the mode of the given data is $2$.

(B) Here,the maximum class frequency is $8$,and the class corresponding to this frequency is $3-5$.
So,the modal class is $3-5$.
Now,
Modal class $= 3-5$,lower limit $(l)$ of modal class $= 3$,class size $(h) = 2$.
Frequency $(f_1)$ of the modal class $= 8$.
Frequency $(f_0)$ of the class preceding the modal class $= 7$.
Frequency $(f_2)$ of the class succeeding the modal class $= 2$.
Now,let us substitute these values in the formula:
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Mode $= 3 + \left( \frac{8 - 7}{2 \times 8 - 7 - 2} \right) \times 2 = 3 + \frac{2}{7} = 3 + 0.286 = 3.286$.
Therefore,the mode of the data above is $3.286$.

(52) To find the mean,we calculate the class marks $(x_i)$ and $f_i x_i$:

Class interval	Number of students $(f_i)$	Class mark $(x_i)$	$f_i x_i$
$10-25$	$2$	$17.5$	$35.0$
$25-40$	$3$	$32.5$	$97.5$
$40-55$	$7$	$47.5$	$332.5$
$55-70$	$6$	$62.5$	$375.0$
$70-85$	$6$	$77.5$	$465.0$
$85-100$	$6$	$92.5$	$555.0$
Total	$\Sigma f_i = 30$		$\Sigma f_i x_i = 1860.0$

Mean $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1860}{30} = 62$.
For the mode,the maximum frequency is $7$,which corresponds to the modal class $40-55$.
Here,$l = 40$,$h = 15$,$f_1 = 7$,$f_0 = 3$,$f_2 = 6$.
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 40 + \left( \frac{7 - 3}{14 - 3 - 6} \right) \times 15 = 40 + \left( \frac{4}{5} \right) \times 15 = 40 + 12 = 52$.
Interpretation: The maximum number of students obtained $52$ marks (mode),while on average,a student obtained $62$ marks (mean).

(N/A) To find the class marks $(x_{i})$,the following relation is used:
$x_{i} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
Taking $30$ as the assumed mean $(a)$,$d_i$ and $f_id_i$ are calculated as follows:

Age (in years)	Number of patients $(f_i)$	Class mark $(x_i)$	$d_i = x_i - 30$	$f_i d_i$
$5$-$15$	$6$	$10$	-$20$	-$120$
$15$-$25$	$11$	$20$	-$10$	-$110$
$25$-$35$	$21$	$30$	$0$	$0$
$35$-$45$	$23$	$40$	$10$	$230$
$45$-$55$	$14$	$50$	$20$	$280$
$55$-$65$	$5$	$60$	$30$	$150$
Total	$80$	-	-	$430$

From the table,we obtain $\Sigma f_{i} = 80$ and $\Sigma f_{i} d_{i} = 430$.
Mean,$\bar{x} = a + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}} = 30 + \frac{430}{80} = 30 + 5.375 = 35.375 \simeq 35.38$.
The mean age of the patients is $35.38 \text{ years}$. This represents that,on average,the age of a patient admitted to the hospital was $35.38 \text{ years}$.
It can be observed that the maximum class frequency is $23$,belonging to the class interval $35-45$.
Modal class $= 35-45$,Lower limit $(l) = 35$,Class size $(h) = 10$,Frequency $(f_1) = 23$,Frequency $(f_0) = 21$,Frequency $(f_2) = 14$.
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 35 + \left( \frac{23 - 21}{2(23) - 21 - 14} \right) \times 10 = 35 + \left( \frac{2}{46 - 35} \right) \times 10 = 35 + \frac{20}{11} = 35 + 1.81 = 36.81$.
The mode is $36.81$. This represents that the age of the maximum number of patients admitted to the hospital was $36.81 \text{ years}$.

(A) From the data given above,it can be observed that the maximum class frequency is $61$,which belongs to the class interval $60-80$.
Therefore,the modal class is $60-80$.
Lower class limit $(l)$ of the modal class $= 60$.
Frequency $(f_1)$ of the modal class $= 61$.
Frequency $(f_0)$ of the class preceding the modal class $= 52$.
Frequency $(f_2)$ of the class succeeding the modal class $= 38$.
Class size $(h) = 20$.
Using the formula for mode:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$= 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20$
$= 60 + \left( \frac{9}{122 - 90} \right) \times 20$
$= 60 + \left( \frac{9}{32} \right) \times 20$
$= 60 + \frac{180}{32} = 60 + 5.625 = 65.625$.
Therefore,the modal lifetime of the electrical components is $65.625 \text{ hours}$.

(N/A) It can be observed from the given data that the maximum class frequency is $40$,belonging to $1500-2000$ interval.
Therefore,modal class $= 1500-2000$.
Lower limit $(l)$ of modal class $= 1500$.
Frequency $(f_1)$ of modal class $= 40$.
Frequency $(f_0)$ of class preceding modal class $= 24$.
Frequency $(f_2)$ of class succeeding modal class $= 33$.
Class size $(h) = 500$.
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$= 1500 + \left( \frac{40 - 24}{2(40) - 24 - 33} \right) \times 500$
$= 1500 + \left( \frac{16}{80 - 57} \right) \times 500 = 1500 + \frac{8000}{23} \approx 1500 + 347.83 = 1847.83$.
Thus,the modal monthly expenditure is Rs. $1847.83$.
To find the mean,we use the step-deviation method:
Mean $(\bar{x}) = a + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right)$
Using $a = 2750$ and $h = 500$,we calculate $\sum f_i u_i = -35$ and $\sum f_i = 200$.
Mean $= 2750 + 500 \times \left( \frac{-35}{200} \right) = 2750 - 87.5 = 2662.5$.
Thus,the mean monthly expenditure is Rs. $2662.5$.

(N/A) It can be observed from the given data that the maximum class frequency is $10$ belonging to class interval $30-35$.
Therefore,modal class $= 30-35$.
Class size $(h) = 5$.
Lower limit $(l)$ of modal class $= 30$.
Frequency $(f_1)$ of modal class $= 10$.
Frequency $(f_0)$ of class preceding modal class $= 9$.
Frequency $(f_2)$ of class succeeding modal class $= 3$.
Mode $= l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 30 + \left(\frac{10 - 9}{2(10) - 9 - 3}\right) \times 5 = 30 + \left(\frac{1}{20 - 12}\right) \times 5 = 30 + \frac{5}{8} = 30.625$.
Mode $\approx 30.6$.
This represents that most of the states/$U$.$T$. have a teacher-student ratio of approximately $30.6$.
To find the mean,we use the step-deviation method.
Mean $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$.
Using assumed mean $a = 32.5$ and $h = 5$,we calculate $\sum f_i u_i = -23$ and $\sum f_i = 35$.
Mean $\bar{x} = 32.5 + \left(\frac{-23}{35}\right) \times 5 = 32.5 - \frac{23}{7} = 32.5 - 3.2857 \approx 29.21$.
Therefore,the mean of the data is approximately $29.2$.

(D) From the given data,the maximum class frequency is $18$,which corresponds to the class interval $4000-5000$.
Therefore,the modal class is $4000-5000$.
Lower limit $(l)$ of the modal class $= 4000$.
Frequency $(f_1)$ of the modal class $= 18$.
Frequency $(f_0)$ of the class preceding the modal class $= 4$.
Frequency $(f_2)$ of the class succeeding the modal class $= 9$.
Class size $(h) = 1000$.
The formula for mode is:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 4000 + \left( \frac{18 - 4}{2(18) - 4 - 9} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{36 - 13} \right) \times 1000$
$\text{Mode} = 4000 + \left( \frac{14}{23} \right) \times 1000$
$\text{Mode} = 4000 + \frac{14000}{23}$
$\text{Mode} = 4000 + 608.695...$
$\text{Mode} \approx 4608.7$
Therefore,the mode of the given data is $4608.7$ runs.

(A) From the given data, it can be observed that the maximum class frequency is $20,$ which belongs to the $40-50$ class interval.
Therefore, the modal class $= 40-50$.
Lower limit $(l)$ of the modal class $= 40$.
Frequency $(f_1)$ of the modal class $= 20$.
Frequency $(f_0)$ of the class preceding the modal class $= 12$.
Frequency $(f_2)$ of the class succeeding the modal class $= 11$.
Class size $(h) = 10$.
Using the mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$.
$\text{Mode} = 40 + \left( \frac{20 - 12}{2(20) - 12 - 11} \right) \times 10$.
$\text{Mode} = 40 + \left( \frac{8}{40 - 23} \right) \times 10$.
$\text{Mode} = 40 + \left( \frac{8}{17} \right) \times 10$.
$\text{Mode} = 40 + \frac{80}{17}$.
$\text{Mode} = 40 + 4.705 \approx 44.7$.
Therefore, the mode of this data is $44.7$ cars.

(B) To calculate the median height,we first convert the cumulative frequency distribution into a standard frequency distribution table.

Class intervals	Frequency $(f)$	Cumulative frequency $(cf)$
Below $140$	$4$	$4$
$140-145$	$7$	$11$
$145-150$	$18$	$29$
$150-155$	$11$	$40$
$155-160$	$6$	$46$
$160-165$	$5$	$51$

Here,$n = 51$. So,$\frac{n}{2} = \frac{51}{2} = 25.5$.
The cumulative frequency just greater than $25.5$ is $29$,which corresponds to the class interval $145-150$. Thus,the median class is $145-150$.
Here,$l = 145$,$cf = 11$,$f = 18$,and $h = 5$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 145 + \left( \frac{25.5 - 11}{18} \right) \times 5$
$\text{Median} = 145 + \left( \frac{14.5}{18} \right) \times 5 = 145 + \frac{72.5}{18} = 145 + 4.027... \approx 149.03 \ cm$.

(X=9, Y=15) First,we construct the cumulative frequency table:

Class intervals	Frequency	Cumulative frequency
$0-100$	$2$	$2$
$100-200$	$5$	$7$
$200-300$	$x$	$7+x$
$300-400$	$12$	$19+x$
$400-500$	$17$	$36+x$
$500-600$	$20$	$56+x$
$600-700$	$y$	$56+x+y$
$700-800$	$9$	$65+x+y$
$800-900$	$7$	$72+x+y$
$900-1000$	$4$	$76+x+y$

Given that the total frequency $n = 100$,we have:
$76 + x + y = 100 \implies x + y = 24$ ........... $(1)$
The median is $525$,which lies in the class $500-600$. Thus,$l = 500$,$f = 20$,$cf = 36 + x$,and $h = 100$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x$
$x = 9$
Substituting $x = 9$ into equation $(1)$:
$9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.

(N/A) $1$. Mean: Using the step deviation method,$\bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$. With $a = 135, h = 20, \sum f_i u_i = 7, \sum f_i = 68$,we get $\bar{x} = 135 + 20 \left( \frac{7}{68} \right) = 135 + 2.06 = 137.06$.
$2$. Mode: Modal class is $125-145$ $(f_1=20, f_0=13, f_2=14)$. Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 125 + \left( \frac{20 - 13}{40 - 13 - 14} \right) \times 20 = 125 + \left( \frac{7}{13} \right) \times 20 = 125 + 10.77 = 135.77$.
$3$. Median: $n=68, n/2=34$. Cumulative frequency table shows median class is $125-145$. Median $= l + \left( \frac{n/2 - cf}{f} \right) \times h = 125 + \left( \frac{34 - 22}{20} \right) \times 20 = 125 + 12 = 137$. Comparing them,Mean $\approx$ Median $\approx$ Mode.

(A) The cumulative frequency for the given data is calculated as follows:

Class interval	Frequency	Cumulative frequency
$0-10$	$5$	$5$
$10-20$	$x$	$5+x$
$20-30$	$20$	$25+x$
$30-40$	$15$	$40+x$
$40-50$	$y$	$40+x+y$
$50-60$	$5$	$45+x+y$

From the table,the total frequency $n = 60$.
Thus,$45+x+y = 60 \implies x+y = 15 \dots (1)$.
The median is $28.5$,which lies in the class interval $20-30$.
Therefore,the median class is $20-30$.
Lower limit $(l) = 20$,frequency $(f) = 20$,cumulative frequency $(cf) = 5+x$,and class size $(h) = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$.
$28.5 = 20 + \left( \frac{30 - (5+x)}{20} \right) \times 10$.
$8.5 = \frac{25-x}{2}$.
$17 = 25 - x \implies x = 8$.
Substituting $x=8$ into equation $(1)$: $8+y = 15 \implies y = 7$.
Thus,$x = 8$ and $y = 7$.

(B) The given data is in the form of 'less than' cumulative frequency distribution. We first convert it into class intervals with their respective frequencies.

Age (in years)	Frequency $(f_i)$	Cumulative frequency $(cf)$
$18-20$	$2$	$2$
$20-25$	$6-2=4$	$6$
$25-30$	$24-6=18$	$24$
$30-35$	$45-24=21$	$45$
$35-40$	$78-45=33$	$78$
$40-45$	$89-78=11$	$89$
$45-50$	$92-89=3$	$92$
$50-55$	$98-92=6$	$98$
$55-60$	$100-98=2$	$100$

Here,$n = 100$,so $\frac{n}{2} = 50$.
The cumulative frequency just greater than $50$ is $78$,which corresponds to the class interval $35-40$.
Thus,the median class is $35-40$.
Lower limit $(l)$ $= 35$,class size $(h)$ $= 5$,frequency of median class $(f)$ $= 33$,and cumulative frequency of the class preceding the median class $(cf)$ $= 45$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 35 + \left( \frac{50 - 45}{33} \right) \times 5$
$\text{Median} = 35 + \left( \frac{5}{33} \right) \times 5 = 35 + \frac{25}{33} \approx 35 + 0.76 = 35.76$.
Therefore,the median age is $35.76$ years.

(C) The given data does not have continuous class intervals. The difference between the upper limit of one class and the lower limit of the next class is $127 - 126 = 1$. Therefore,we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of each class to make them continuous.
Continuous class intervals and cumulative frequencies are as follows:

Length (in mm)	Frequency $(f_i)$	Cumulative Frequency $(cf)$
$117.5-126.5$	$3$	$3$
$126.5-135.5$	$5$	$8$
$135.5-144.5$	$9$	$17$
$144.5-153.5$	$12$	$29$
$153.5-162.5$	$5$	$34$
$162.5-171.5$	$4$	$38$
$171.5-180.5$	$2$	$40$

Here,$n = 40$,so $\frac{n}{2} = 20$. The cumulative frequency just greater than $20$ is $29$,which corresponds to the class interval $144.5-153.5$.
Median class $= 144.5-153.5$,lower limit $(l)$ $= 144.5$,class size $(h)$ $= 9$,frequency $(f)$ $= 12$,and cumulative frequency of the preceding class $(cf)$ $= 17$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 144.5 + \left( \frac{20 - 17}{12} \right) \times 9 = 144.5 + \left( \frac{3}{12} \right) \times 9 = 144.5 + 2.25 = 146.75$.
Thus,the median length of the leaves is $146.75 \text{ mm}$.

(D) The cumulative frequencies with their respective class intervals are as follows:

Life time	Number of lamps $(f_i)$	Cumulative frequency $(cf)$
$1500-2000$	$14$	$14$
$2000-2500$	$56$	$70$
$2500-3000$	$60$	$130$
$3000-3500$	$86$	$216$
$3500-4000$	$74$	$290$
$4000-4500$	$62$	$352$
$4500-5000$	$48$	$400$

Here,$n = 400$,so $\frac{n}{2} = 200$.
The cumulative frequency just greater than $200$ is $216$,which corresponds to the class interval $3000-3500$.
Thus,the median class is $3000-3500$.
Lower limit $(l)$ of the median class = $3000$.
Frequency $(f)$ of the median class = $86$.
Cumulative frequency $(cf)$ of the class preceding the median class = $130$.
Class size $(h)$ = $500$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 3000 + \left( \frac{200 - 130}{86} \right) \times 500$
$\text{Median} = 3000 + \left( \frac{70}{86} \right) \times 500$
$\text{Median} = 3000 + \frac{35000}{86} \approx 3000 + 406.98 = 3406.98$.
Therefore,the median lifetime of a lamp is $3406.98 \text{ hours}$.

(N/A) The cumulative frequencies are as follows:

Number of letters	Frequency $(f)$	Cumulative frequency $(cf)$
$1-4$	$6$	$6$
$4-7$	$30$	$36$
$7-10$	$40$	$76$
$10-13$	$16$	$92$
$13-16$	$4$	$96$
$16-19$	$4$	$100$

$1$. Median: $n=100$,so $n/2 = 50$. The cumulative frequency just greater than $50$ is $76$,which corresponds to the class $7-10$.
Median $= l + [(\frac{n}{2} - cf) / f] \times h = 7 + [(50 - 36) / 40] \times 3 = 7 + (14/40) \times 3 = 7 + 1.05 = 8.05$.
$2$. Mean: Using assumed mean $a = 11.5$ and $h = 3$,we calculate $x_i$ (midpoints): $2.5, 5.5, 8.5, 11.5, 14.5, 17.5$.
$u_i = (x_i - 11.5)/3$: $-3, -2, -1, 0, 1, 2$.
$f_i u_i$: $-18, -60, -40, 0, 4, 8$. Sum $\sum f_i u_i = -106$.
Mean $= a + h \times (\sum f_i u_i / \sum f_i) = 11.5 + 3 \times (-106/100) = 11.5 - 3.18 = 8.32$.
$3$. Mode: The modal class is $7-10$ (highest frequency $40$).
Mode $= l + [(f_1 - f_0) / (2f_1 - f_0 - f_2)] \times h = 7 + [(40 - 30) / (80 - 30 - 16)] \times 3 = 7 + (10/34) \times 3 = 7 + 0.88 = 7.88$.

(C) The cumulative frequencies with their respective class intervals are as follows:

Weight (in kg)	Frequency $(f_i)$	Cumulative frequency $(cf)$
$40-45$	$2$	$2$
$45-50$	$3$	$5$
$50-55$	$8$	$13$
$55-60$	$6$	$19$
$60-65$	$6$	$25$
$65-70$	$3$	$28$
$70-75$	$2$	$30$

Total number of students $n = 30$.
We calculate $\frac{n}{2} = \frac{30}{2} = 15$.
The cumulative frequency just greater than $15$ is $19$,which corresponds to the class interval $55-60$.
Thus,the median class is $55-60$.
Lower limit of the median class $(l)$ = $55$.
Frequency of the median class $(f)$ = $6$.
Cumulative frequency of the class preceding the median class $(cf)$ = $13$.
Class size $(h)$ = $5$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$.
$\text{Median} = 55 + \left( \frac{15 - 13}{6} \right) \times 5$.
$\text{Median} = 55 + \left( \frac{2}{6} \right) \times 5 = 55 + \frac{10}{6} = 55 + 1.67 = 56.67$.
Therefore,the median weight of the students is $56.67 \text{ kg}$.

(17.5) To draw the 'more than' ogive,we plot the points $(5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7),$ and $(35, 3)$ on a graph where the horizontal axis represents the lower limits of profit and the vertical axis represents the cumulative frequency. We join these points with a smooth curve.
Next,we convert the given data into a frequency distribution table:

Classes	$5-10$	$10-15$	$15-20$	$20-25$	$25-30$	$30-35$	$35-40$
No. of shops	$2$	$12$	$2$	$4$	$3$	$4$	$3$
Cumulative frequency	$2$	$14$	$16$	$20$	$23$	$27$	$30$

To draw the 'less than' ogive,we plot the points $(10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27),$ and $(40, 30)$ on the same axes.
The abscissa of the point of intersection of the two ogives gives the median. From the graph,the point of intersection is at $x = 17.5$. Therefore,the median profit is $Rs. 17.5$ lakhs.

(N/A) The frequency distribution table of less than type is as follows:

Daily income (in Rs.) (upper class limits)	Cumulative frequency
Less than $120$	$12$
Less than $140$	$12 + 14 = 26$
Less than $160$	$26 + 8 = 34$
Less than $180$	$34 + 6 = 40$
Less than $200$	$40 + 10 = 50$

Taking upper class limits of class intervals on the $x$-axis and their respective cumulative frequencies on the $y$-axis,the ogive is drawn by plotting the points $(120, 12), (140, 26), (160, 34), (180, 40),$ and $(200, 50)$ and joining them with a smooth curve.

(N/A) The given cumulative frequency distributions of less than type are:

Weight (in kg) (upper class limits)	Number of students (cumulative frequency)
Less than $38$	$0$
Less than $40$	$3$
Less than $42$	$5$
Less than $44$	$9$
Less than $46$	$14$
Less than $48$	$28$
Less than $50$	$32$
Less than $52$	$35$

Taking upper class limits on the $x$-axis and their respective cumulative frequencies on the $y$-axis,the ogive is drawn as shown in the figure.
Here,$n = 35$.
So,$\frac{n}{2} = 17.5$.
Mark the point $A$ on the curve whose ordinate is $17.5$. The $x$-coordinate corresponding to this point is $46.5$. Therefore,the median of this data is $46.5$.
To verify using the formula,we first determine the class intervals and their frequencies:

Weight (in kg)	Frequency $(f)$	Cumulative frequency $(cf)$
$38-40$	$3-0=3$	$3$
$40-42$	$5-3=2$	$5$
$42-44$	$9-5=4$	$9$
$44-46$	$14-9=5$	$14$
$46-48$	$28-14=14$	$28$
$48-50$	$32-28=4$	$32$
$50-52$	$35-32=3$	$35$

The cumulative frequency just greater than $\frac{n}{2} = 17.5$ is $28$,which belongs to the class interval $46-48$.
Median class $= 46-48$.
Lower class limit $(l) = 46$.
Frequency $(f) = 14$.
Cumulative frequency of the class preceding the median class $(cf) = 14$.
Class size $(h) = 2$.
Using the median formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$= 46 + \left( \frac{17.5 - 14}{14} \right) \times 2$
$= 46 + \left( \frac{3.5}{14} \right) \times 2$
$= 46 + \frac{7}{14} = 46 + 0.5 = 46.5$.
The result is verified.

(N/A) The cumulative frequency distribution of 'more than type' can be obtained as follows:

Production yield (lower class limits)	Cumulative frequency
More than or equal to $50$	$100$
More than or equal to $55$	$100 - 2 = 98$
More than or equal to $60$	$98 - 8 = 90$
More than or equal to $65$	$90 - 12 = 78$
More than or equal to $70$	$78 - 24 = 54$
More than or equal to $75$	$54 - 38 = 16$

By plotting the points $(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16)$ on a graph paper,taking the lower class limits on the $x$-axis and their respective cumulative frequencies on the $y$-axis,we obtain the 'more than type' ogive.