Mix Examples - Statistics Questions in English

(A) cumulative frequency table is a statistical tool used to organize data by summing the frequencies of each class interval up to a certain point.
This table is specifically required to calculate the median of a grouped frequency distribution.
The formula for the median is $Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,where $cf$ represents the cumulative frequency of the class preceding the median class.
While mean and mode can be calculated using simple frequency tables,the cumulative frequency is essential for the median.

(B) The given data is a 'more than' cumulative frequency distribution.
To find the number of families in the income range $16000-19000$,we subtract the number of families having income more than $19000$ from the number of families having income more than $16000$.
Number of families with income $> 16000 = 69$.
Number of families with income $> 19000 = 50$.
Number of families in the range $16000-19000 = 69 - 50 = 19$.
Therefore,the correct option is $B$.

(C) $1$. Modal Class: The modal class is the class interval with the highest frequency. Here,the highest frequency is $15$,which corresponds to the class interval $150-155$. Thus,the lower limit of the modal class is $150$.
$2$. Median Class: To find the median class,we calculate the cumulative frequency $(cf)$:
- $150-155$: $cf = 15$
- $155-160$: $cf = 15 + 13 = 28$
- $160-165$: $cf = 28 + 10 = 38$
- $165-170$: $cf = 38 + 8 = 46$
- $170-175$: $cf = 46 + 9 = 55$
- $175-180$: $cf = 55 + 5 = 60$
The total number of students $N = 60$. So,$N/2 = 30$. The cumulative frequency just greater than $30$ is $38$,which corresponds to the class interval $160-165$. Thus,the upper limit of the median class is $165$.
$3$. Final Calculation: The sum of the lower limit of the modal class $(150)$ and the upper limit of the median class $(165)$ is $150 + 165 = 315$.

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

Class	Frequency $(f_i)$	Midpoint $(x_i)$	$d_i = x_i - A$	$u_i = \frac{x_i - A}{c}$	$f_i x_i$	$f_i d_i$	$f_i u_i$
$50-70$	$10$	$60$	$-20$	$-1$	$600$	$-200$	$-10$
$70-90$	$18$	$80 = A$	$0$	$0$	$1440$	$0$	$0$
$90-110$	$7$	$100$	$20$	$1$	$700$	$140$	$7$
$110-130$	$6$	$120$	$40$	$2$	$720$	$240$	$12$
$130-150$	$5$	$140$	$60$	$3$	$700$	$300$	$15$
$150-170$	$4$	$160$	$80$	$4$	$640$	$320$	$16$
Total	$\Sigma f_i = 50$	-	-	-	$\Sigma f_i x_i = 4800$	$\Sigma f_i d_i = 800$	$\Sigma f_i u_i = 40$

Here,$A = 80$ and $c = 20$.
$1$. Direct Method: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4800}{50} = 96$.
$2$. Assumed Mean Method: $\bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} = 80 + \frac{800}{50} = 80 + 16 = 96$.
$3$. Step Deviation Method: $\bar{x} = A + \left( \frac{\Sigma f_i u_i}{\Sigma f_i} \right) \times c = 80 + \left( \frac{40}{50} \right) \times 20 = 80 + 16 = 96$.
Thus,the mean of the data is $96$.

(A) To find the mean,we use the step-deviation method with assumed mean $A = 549.5$ and class size $c = 100$.

Class	Frequency $(f_{i})$	Midpoint $(x_{i})$	$u_{i} = \frac{x_{i} - A}{c}$	$f_{i} u_{i}$
$200-299$	$3$	$249.5$	$-3$	$-9$
$300-399$	$61$	$349.5$	$-2$	$-122$
$400-499$	$118$	$449.5$	$-1$	$-118$
$500-599$	$139$	$549.5 (A)$	$0$	$0$
$600-699$	$126$	$649.5$	$1$	$126$
$700-799$	$151$	$749.5$	$2$	$302$
$800-899$	$2$	$849.5$	$3$	$6$
Total	$\Sigma f_{i} = 600$	-	-	$\Sigma f_{i} u_{i} = 185$

The formula for the mean is $\bar{x} = A + \left( \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \right) \times c$.
Substituting the values:
$\bar{x} = 549.5 + \left( \frac{185}{600} \right) \times 100$
$\bar{x} = 549.5 + \frac{185}{6}$
$\bar{x} = 549.5 + 30.833...$
$\bar{x} \approx 580.33$
Thus,the mean of the data is $580.33$.

(B) To find the missing frequency $f$,we use the step-deviation method for mean: $\bar{x} = A + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times c$.

Class	Frequency $(f_i)$	Midpoint $(x_i)$	$u_i = \frac{x_i - A}{c}$	$f_i u_i$
$0-10$	$8$	$5$	$-4$	$-32$
$10-20$	$4$	$15$	$-3$	$-12$
$20-30$	$20$	$25$	$-2$	$-40$
$30-40$	$45$	$35$	$-1$	$-45$
$40-50$	$64$	$45=A$	$0$	$0$
$50-60$	$32$	$55$	$1$	$32$
$60-70$	$f$	$65$	$2$	$2f$
$70-80$	$8$	$75$	$3$	$24$
$80-90$	$2$	$85$	$4$	$8$
$90-100$	$2$	$95$	$5$	$10$
Total	$\sum f_i = 185 + f$	-	-	$\sum f_i u_i = 2f - 55$

Here,$A = 45$ and $c = 10$.
Given $\bar{x} = 43.75$.
$43.75 = 45 + \frac{2f - 55}{185 + f} \times 10$
$-1.25 = \frac{10(2f - 55)}{185 + f}$
$-1.25(185 + f) = 20f - 550$
$-231.25 - 1.25f = 20f - 550$
$550 - 231.25 = 20f + 1.25f$
$318.75 = 21.25f$
$f = \frac{318.75}{21.25} = 15$.
Thus,the missing frequency $f = 15$.

(C) The formula for the mean of grouped data using the assumed mean method is $\bar{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$.
In this formula,$a$ is the assumed mean.
The term $d_{i}$ represents the deviation of the class marks $(x_{i})$ from the assumed mean $(a)$.
Therefore,$d_{i} = x_{i} - a$,where $x_{i}$ are the mid-points of the classes.

(D) In computing the mean of grouped data,we assume that each observation in a class is equal to its class mark. Therefore,the frequencies are assumed to be centred at the class marks of the classes.

(A) The mean $\bar{x}$ of grouped data is given by the formula $\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$,where $n = \sum f_{i}$.
This can be rewritten as $\sum f_{i} x_{i} = n \bar{x}$.
We need to evaluate the sum $\sum (f_{i} x_{i} - f_{i} \bar{x})$.
Using the properties of summation,we get $\sum f_{i} x_{i} - \sum f_{i} \bar{x}$.
Since $\bar{x}$ is a constant,we can take it out of the summation: $\sum f_{i} x_{i} - \bar{x} \sum f_{i}$.
Substituting $\sum f_{i} x_{i} = n \bar{x}$ and $\sum f_{i} = n$,we get $n \bar{x} - \bar{x} (n) = n \bar{x} - n \bar{x} = 0$.

(B) The given formula $\bar{x} = a + h \left( \frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}} \right)$ is the step-deviation method formula for calculating the mean of a grouped frequency distribution.
In this formula,$a$ represents the assumed mean,$h$ represents the class size,$x_{i}$ represents the class mark,and $u_{i}$ is defined as the deviation divided by the class size.
Therefore,the expression for $u_{i}$ is $u_{i} = \frac{x_{i} - a}{h}$.

(C) The cumulative frequency curves are also known as ogives. The 'less than' type ogive and 'more than' type ogive are plotted on the same graph. The point where these two curves intersect represents the median of the grouped data. The $x$-coordinate (abscissa) of this intersection point corresponds to the median value of the data set.

(D) To find the median class and modal class,we first prepare the cumulative frequency table:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-5$	$10$	$10$
$5-10$	$15$	$25$
$10-15$	$12$	$37$
$15-20$	$20$	$57$
$20-25$	$9$	$66$

$1$. Total frequency $N = 66$. So,$\frac{N}{2} = \frac{66}{2} = 33$.
$2$. The cumulative frequency just greater than $33$ is $37$,which corresponds to the class interval $10-15$. Thus,the median class is $10-15$,and its lower limit is $10$.
$3$. The highest frequency is $20$,which corresponds to the class interval $15-20$. Thus,the modal class is $15-20$,and its lower limit is $15$.
$4$. The sum of the lower limits is $10 + 15 = 25$.

(A) To find the median class,we first convert the discontinuous class intervals into continuous ones by adjusting the boundaries. The difference between the upper limit of one class and the lower limit of the next is $1$. Thus,we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit.

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$-0.5-5.5$	$13$	$13$
$5.5-11.5$	$10$	$23$
$11.5-17.5$	$15$	$38$
$17.5-23.5$	$8$	$46$
$23.5-29.5$	$11$	$57$

The total frequency $N = 57$. The median position is $N/2 = 57/2 = 28.5$. The cumulative frequency just greater than $28.5$ is $38$,which corresponds to the class interval $11.5-17.5$. Therefore,the median class is $11.5-17.5$. The upper limit of this class is $17.5$.

(B) To find the modal class,we first convert the cumulative frequency distribution into a standard frequency distribution table:

Class Interval	Frequency
$0-10$	$3$
$10-20$	$12 - 3 = 9$
$20-30$	$27 - 12 = 15$
$30-40$	$57 - 27 = 30$
$40-50$	$75 - 57 = 18$
$50-60$	$80 - 75 = 5$

The modal class is the class interval with the highest frequency.
From the table,the highest frequency is $30$,which corresponds to the class interval $30-40$.
Therefore,the modal class is $30-40$.

(C) To find the median class and modal class,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$65-85$	$4$	$4$
$85-105$	$5$	$9$
$105-125$	$13$	$22$
$125-145$	$20$	$42$
$145-165$	$14$	$56$
$165-185$	$7$	$63$
$185-205$	$4$	$67$

Total frequency $N = 67$. Thus,$\frac{N}{2} = \frac{67}{2} = 33.5$.
The cumulative frequency just greater than $33.5$ is $42$,which corresponds to the class interval $125-145$. Therefore,the median class is $125-145$. The upper limit of the median class is $145$.
The highest frequency is $20$,which corresponds to the class interval $125-145$. Therefore,the modal class is $125-145$. The lower limit of the modal class is $125$.
Required difference = (Upper limit of median class) - (Lower limit of modal class) = $145 - 125 = 20$.

(D) To find the number of athletes who completed the race in less than $14.6\, s$,we need to sum the frequencies of all classes that are less than $14.6$.
The classes less than $14.6$ are:
$13.8-14.0$ (Frequency = $2$)
$14.0-14.2$ (Frequency = $4$)
$14.2-14.4$ (Frequency = $5$)
$14.4-14.6$ (Frequency = $71$)
Total number of athletes $= 2 + 4 + 5 + 71 = 82$.

(A) To find the frequency of the class interval $30-40$, we subtract the number of students who scored more than or equal to $40$ from the number of students who scored more than or equal to $30$.
Frequency of class $30-40 = (\text{Number of students } \ge 30) - (\text{Number of students } \ge 40)$
Frequency of class $30-40 = 51 - 48 = 3$.
Thus, the frequency of the class $30-40$ is $3$.

(B) The statement is not true.
When we calculate the mean of grouped data,we assume that the frequency of each class is centered at the mid-point of that class.
In reality,the individual observations within a class are rarely distributed exactly at the mid-point.
Therefore,the mean calculated from grouped data is an approximation,and it is rarely the same as the mean calculated from the original ungrouped data.

(B) No,it is not correct to say that an ogive is a graphical representation of a frequency distribution.
An ogive is specifically a graphical representation of a $cumulative$ frequency distribution.
$A$ frequency distribution is typically represented by a histogram,frequency polygon,or frequency curve.
Therefore,an ogive is used to represent the cumulative frequency,not the simple frequency distribution.

(N/A) No,this is not a correct statement. The median of ungrouped data and the median calculated from grouped data are not always the same. This is because the formula used for calculating the median of grouped data,$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,is based on the assumption that the observations within each class interval are uniformly distributed. In reality,the actual distribution of data within classes may not be uniform,leading to a slight difference between the two values.

(N/A) No,the statement is incorrect. While it is common practice to choose the assumed mean $a$ as one of the class marks (mid-points) to simplify calculations,it is not a mathematical requirement. The assumed mean $a$ can be any arbitrary value chosen from the data set or even outside it,provided it helps in simplifying the calculation of deviations $d_i = x_i - a$.

(N/A) No,it is not true to say that the mean,mode,and median of grouped data will always be different.
These three measures of central tendency can be the same depending on the nature of the data.
For example,in a perfectly symmetrical distribution (such as a normal distribution),the mean,median,and mode are equal.
Therefore,the values of these three measures depend entirely on the distribution of the data.

(N/A) No,the median class and the modal class of grouped data are not always different. They can be the same depending on the distribution of the data.
Justification:
$1$. The modal class is the class interval with the highest frequency.
$2$. The median class is the class interval where the cumulative frequency reaches $N/2$,where $N$ is the total number of observations.
$3$. In many frequency distributions,the class with the highest frequency often contains the middle observation. For example,in a symmetric distribution,the mean,median,and mode often coincide,and the median class and modal class frequently overlap or are identical.

(N/A) To construct the cumulative frequency distribution,we add the frequency of each class to the sum of the frequencies of all preceding classes.

Class	Frequency	Cumulative frequency
$12.5-17.5$	$2$	$2$
$17.5-22.5$	$22$	$2 + 22 = 24$
$22.5-27.5$	$19$	$24 + 19 = 43$
$27.5-32.5$	$14$	$43 + 14 = 57$
$32.5-37.5$	$13$	$57 + 13 = 70$

(A) To find the mean daily wages,we use the direct method formula: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
First,we calculate the class mark $(x_i)$ for each interval using the formula: $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$.

Daily wages (in $Rs.$)	Class mark $(x_i)$	Number of workers $(f_i)$	$f_i x_i$
$100-120$	$110$	$10$	$1100$
$120-140$	$130$	$15$	$1950$
$140-160$	$150$	$20$	$3000$
$160-180$	$170$	$22$	$3740$
$180-200$	$190$	$18$	$3420$
$200-220$	$210$	$12$	$2520$
$220-240$	$230$	$13$	$2990$

Sum of frequencies $\sum f_i = 10 + 15 + 20 + 22 + 18 + 12 + 13 = 110$.
Sum of products $\sum f_i x_i = 1100 + 1950 + 3000 + 3740 + 3420 + 2520 + 2990 = 18720$.
Mean $\bar{x} = \frac{18720}{110} \approx 170.18$.
Thus,the mean daily wage is $Rs. 170.18$.

(B)

Marks (Class)	Frequency $(f)$	Cumulative Frequency $(cf)$
$30-35$	$14$	$14$
$35-40$	$16$	$30$
$40-45$	$18$	$48$
$45-50$	$23$	$71$
$50-55$	$18$	$89$
$55-60$	$8$	$97$
$60-65$	$3$	$100$

Here,$n = 100$.
Therefore,$\frac{n}{2} = \frac{100}{2} = 50$.
The cumulative frequency just greater than $50$ is $71$,which corresponds to the class interval $45-50$. Thus,the median class is $45-50$.
$l$ (lower limit of median class) $= 45$
$cf$ (cumulative frequency of class preceding median class) $= 48$
$f$ (frequency of median class) $= 23$
$h$ (class size) $= 5$
Median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$\text{Median} = 45 + \left( \frac{50 - 48}{23} \right) \times 5$
$\text{Median} = 45 + \left( \frac{2}{23} \right) \times 5$
$\text{Median} = 45 + \frac{10}{23} \approx 45 + 0.4347 \approx 45.43$
Rounding to one decimal place,the median percentage of marks is $45.4$.

(C) Here,the maximum class frequency is $80,$ and the class corresponding to this frequency is $5-7.$
So,the modal class is $5-7.$
$l$ (lower limit of modal class) $= 5$
$f_{1}$ (frequency of the modal class) $= 80$
$f_{0}$ (frequency of the class preceding the modal class) $= 45$
$f_{2}$ (frequency of the class succeeding the modal class) $= 55$
$h$ (class size) $= 2$
Mode $= l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times h$
Mode $= 5 + \left( \frac{80 - 45}{2(80) - 45 - 55} \right) \times 2$
Mode $= 5 + \left( \frac{35}{160 - 100} \right) \times 2$
Mode $= 5 + \left( \frac{35}{60} \right) \times 2 = 5 + \frac{35}{30}$
Mode $= 5 + 1.166... \approx 6.17$ (Rounding to $6.2$ as per options).
Hence,the modal agricultural holdings of the village is $6.2$ hectares.

(D) To find the mean,we first calculate the class mark $(x_i)$ for each class interval using the formula: $x_i = \frac{\text{Upper limit} + \text{Lower limit}}{2}$.

Class	Class mark $(x_i)$	Frequency $(f_i)$	$f_i x_i$
$1-3$	$2$	$9$	$18$
$3-5$	$4$	$22$	$88$
$5-7$	$6$	$27$	$162$
$7-10$	$8.5$	$17$	$144.5$
Total	-	$\Sigma f_i = 75$	$\Sigma f_i x_i = 412.5$

The mean $(\bar{x})$ is calculated as:
$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{412.5}{75} = 5.5$.
Thus,the mean of the given distribution is $5.5$.

(A) To find the mean,we first calculate the class mark $(x_i)$ for each class interval using the formula: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.

Marks	Class marks $(x_i)$	Frequency $(f_i)$	$f_i x_i$
$10-20$	$15$	$2$	$30$
$20-30$	$25$	$4$	$100$
$30-40$	$35$	$7$	$245$
$40-50$	$45$	$6$	$270$
$50-60$	$55$	$1$	$55$
Total	-	$\Sigma f_i = 20$	$\Sigma f_i x_i = 700$

The mean $(\bar{x})$ is calculated as: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{700}{20} = 35$.
Thus,the mean score of the $20$ students is $35$.

(B) Since the given data is not continuous,we first convert it into continuous class intervals by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class.
Now,we calculate the class mark $x_{i}$ for each class and then proceed as follows:

Class	Class mark $(x_{i})$	Frequency $(f_{i})$	$f_{i} x_{i}$
$3.5-7.5$	$5.5$	$5$	$27.5$
$7.5-11.5$	$9.5$	$4$	$38$
$11.5-15.5$	$13.5$	$9$	$121.5$
$15.5-19.5$	$17.5$	$10$	$175$
Total	-	$\Sigma f_{i} = 28$	$\Sigma f_{i} x_{i} = 362$

Therefore,the mean $\bar{x} = \frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}} = \frac{362}{28} \approx 12.93$.
Hence,the mean of the given data is $12.93$.

(C) To find the mean,we first make the class intervals continuous by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit. Then we calculate the mid-value $(x_i)$ for each class.

Class Interval	Mid Value $(x_i)$	Frequency $(f_i)$	$f_i x_i$
$15.5-18.5$	$17$	$1$	$17$
$18.5-21.5$	$20$	$3$	$60$
$21.5-24.5$	$23$	$4$	$92$
$24.5-27.5$	$26$	$9$	$234$
$27.5-30.5$	$29$	$13$	$377$
Total	-	$\Sigma f_i = 30$	$\Sigma f_i x_i = 780$

The mean is calculated using the formula: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$.
$\bar{x} = \frac{780}{30} = 26$.
Thus,the mean number of pages written per day is $26$.

(D) Since the given data is not continuous,we first convert it into continuous classes by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class.
Now,we calculate the class mark $(x_i)$ for each class and use the step-deviation method:

Income (in $Rs.$)	Class mark $(x_i)$	Frequency $(f_i)$	$u_i = \frac{x_i - a}{h}$	$f_i u_i$
$0.5-200.5$	$100.5$	$14$	$-1$	$-14$
$200.5-400.5$	$300.5 (a)$	$15$	$0$	$0$
$400.5-600.5$	$500.5$	$14$	$1$	$14$
$600.5-800.5$	$700.5$	$7$	$2$	$14$
Total	-	$\Sigma f_i = 50$	-	$\Sigma f_i u_i = 14$

Here,assumed mean $a = 300.5$,class width $h = 200$,and total observations $N = 50$.
Using the step-deviation formula:
Mean $(\bar{x}) = a + h \times \left( \frac{\Sigma f_i u_i}{\Sigma f_i} \right)$
$\bar{x} = 300.5 + 200 \times \left( \frac{14}{50} \right)$
$\bar{x} = 300.5 + 4 \times 14$
$\bar{x} = 300.5 + 56 = 356.5$
Thus,the mean daily income is $Rs. 356.5$.

(A) To find the mean number of seats occupied,we use the assumed mean method. First,we calculate the class mark $(x_i)$ for each class interval.

Number of seats	Class marks $(x_i)$	Frequency $(f_i)$	Deviation $d_i = x_i - a$	$f_i d_i$
$100-104$	$102$	$15$	$-8$	$-120$
$104-108$	$106$	$20$	$-4$	$-80$
$108-112$	$a=110$	$32$	$0$	$0$
$112-116$	$114$	$18$	$4$	$72$
$116-120$	$118$	$15$	$8$	$120$
Total	-	$\Sigma f_i = 100$	-	$\Sigma f_i d_i = -8$

Assumed mean $(a) = 110$.
Using the formula for the mean by assumed mean method:
$\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$
$\bar{x} = 110 + \left(\frac{-8}{100}\right)$
$\bar{x} = 110 - 0.08 = 109.92$
Thus,the mean number of seats occupied is $109.92$.

(B) To find the mean weight,we use the assumed mean method. First,we determine the class marks $(x_i)$ for each class interval.

Weight (in $kg$)	Number of wrestlers $(f_i)$	Class Marks $(x_i)$	Deviations $(d_i = x_i - a)$	$f_i d_i$
$100-110$	$4$	$105$	$-20$	$-80$
$110-120$	$14$	$115$	$-10$	$-140$
$120-130$	$21$	$a = 125$	$0$	$0$
$130-140$	$8$	$135$	$10$	$80$
$140-150$	$3$	$145$	$20$	$60$
Total	$\Sigma f_i = 50$	-	-	$\Sigma f_i d_i = -80$

Let the assumed mean $(a) = 125$.
Using the formula for the mean by the assumed mean method:
Mean $(\bar{x}) = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$
$= 125 + \frac{-80}{50}$
$= 125 - 1.6$
$= 123.4 \ kg$.

(A)

Mileage $(km/l)$	Class marks $(x_i)$	Number of cars $(f_i)$	$f_i x_i$
$10-12$	$11$	$7$	$77$
$12-14$	$13$	$12$	$156$
$14-16$	$15$	$18$	$270$
$16-18$	$17$	$13$	$221$
Total	-	$\Sigma f_i = 50$	$\Sigma f_i x_i = 724$

Here,$\Sigma f_i = 50$ and $\Sigma f_i x_i = 724$.
Mean $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{724}{50} = 14.48$.
Hence,the mean mileage is $14.48 \, km/l$.
No,$I$ do not agree with the manufacturer's claim,as the calculated mean mileage $(14.48 \, km/l)$ is significantly less than the claimed $16 \, km/l$.

(N/A) To construct a cumulative frequency distribution (less than type),we add the frequencies of all classes preceding the upper limit of the current class.

Weight (in $kg$)	Cumulative Frequency
Less than $45$	$4$
Less than $50$	$4 + 4 = 8$
Less than $55$	$8 + 13 = 21$
Less than $60$	$21 + 5 = 26$
Less than $65$	$26 + 6 = 32$
Less than $70$	$32 + 5 = 37$
Less than $75$	$37 + 2 = 39$
Less than $80$	$39 + 1 = 40$

(N/A) To construct the frequency distribution table,we subtract the cumulative frequency of the previous class from the current class to find the frequency of each class interval.

Class interval	Number of students (Frequency)
$0-10$	$10$
$10-20$	$50 - 10 = 40$
$20-30$	$130 - 50 = 80$
$30-40$	$270 - 130 = 140$
$40-50$	$440 - 270 = 170$
$50-60$	$570 - 440 = 130$
$60-70$	$670 - 570 = 100$
$70-80$	$740 - 670 = 70$
$80-90$	$780 - 740 = 40$
$90-100$	$800 - 780 = 20$

(N/A) To form the frequency distribution table,we subtract the cumulative frequencies of consecutive intervals to find the frequency of each specific class interval.
Given that the total number of candidates is $34$ (as seen in the 'More than or equal to $0$' category),we calculate the frequency for each class interval as follows:

Class interval	Frequency
$0-10$	$34-32 = 2$
$10-20$	$32-30 = 2$
$20-30$	$30-27 = 3$
$30-40$	$27-23 = 4$
$40-50$	$23-17 = 6$
$50-60$	$17-11 = 6$
$60-70$	$11-6 = 5$
$70-80$	$6-4 = 2$
$80-90$	$4$

(A) To find the unknown entries,we use the definition of cumulative frequency,where each entry is the sum of the frequency of the current class and all previous classes.
$1$. For the first class $(150-155)$,the cumulative frequency is equal to the frequency: $a = 12$.
$2$. For the second class $(155-160)$,the cumulative frequency is $12 + b = 25$. Thus,$b = 25 - 12 = 13$.
$3$. For the third class $(160-165)$,the cumulative frequency is $c = 12 + b + 10 = 12 + 13 + 10 = 35$.
$4$. For the fourth class $(165-170)$,the cumulative frequency is $c + d = 43$. Substituting $c = 35$,we get $35 + d = 43$,so $d = 43 - 35 = 8$.
$5$. For the fifth class $(170-175)$,the cumulative frequency is $43 + e = 48$. Thus,$e = 48 - 43 = 5$.
$6$. For the sixth class $(175-180)$,the cumulative frequency is $f = 48 + 2 = 50$. Thus,$f = 50$.
Therefore,the values are: $a = 12, b = 13, c = 35, d = 8, e = 5, f = 50$.

(N/A) $(i)$ To form a 'less than' type cumulative frequency distribution,we add the frequencies of all classes preceding the current class. Since the data starts from $10-20$,we assume $0$ patients are less than $10$ years old.
$(ii)$ To form a 'more than' type cumulative frequency distribution,we start with the total number of patients $(300)$ and subtract the frequency of each preceding class interval.

Age (in years)	Less than type (Cumulative Frequency)	Age (in years)	More than type (Cumulative Frequency)
Less than $10$	$0$	$10$ or more	$300$
Less than $20$	$60$	$20$ or more	$240$
Less than $30$	$102$	$30$ or more	$198$
Less than $40$	$157$	$40$ or more	$143$
Less than $50$	$227$	$50$ or more	$73$
Less than $60$	$280$	$60$ or more	$20$
Less than $70$	$300$	$70$ or more	$0$

(N/A) To convert the cumulative frequency distribution into a frequency distribution,we subtract the cumulative frequency of the preceding class from the cumulative frequency of the current class.
$1$. For the interval $0-20$,the frequency is $17$.
$2$. For the interval $20-40$,the frequency is $22 - 17 = 5$.
$3$. For the interval $40-60$,the frequency is $29 - 22 = 7$.
$4$. For the interval $60-80$,the frequency is $37 - 29 = 8$.
$5$. For the interval $80-100$,the frequency is $50 - 37 = 13$.
The resulting frequency distribution table is:

Marks	Number of students
$0-20$	$17$
$20-40$	$5$
$40-60$	$7$
$60-80$	$8$
$80-100$	$13$

(B) First,we construct a cumulative frequency table:

Weekly income (in $Rs.$)	Number of families $(f_i)$	Cumulative frequency $(cf)$
$0-1000$	$250$	$250$
$1000-2000$	$190$	$440$
$2000-3000$	$100$	$540$
$3000-4000$	$40$	$580$
$4000-5000$	$15$	$595$
$5000-6000$	$5$	$600$

Given $n = 600$,we find $\frac{n}{2} = \frac{600}{2} = 300$.
The cumulative frequency just greater than $300$ is $440$,which corresponds to the median class $1000-2000$.
Here,lower limit $l = 1000$,frequency $f = 190$,cumulative frequency of the preceding class $cf = 250$,and class size $h = 1000$.
Using the median formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$= 1000 + \left( \frac{300 - 250}{190} \right) \times 1000$
$= 1000 + \left( \frac{50}{190} \right) \times 1000$
$= 1000 + \frac{5000}{19}$
$= 1000 + 263.157... \approx 1263.16$
Rounding to two decimal places,the median income is $Rs. 1263.15$.

(C) Given total number of players $n = 33$.
$\therefore \frac{n}{2} = \frac{33}{2} = 16.5$.
To find the median class,we look at the cumulative frequencies:
- $85-100$: $11$
- $100-115$: $11 + 9 = 20$
Since $16.5$ lies in the cumulative frequency $20$,the median class is $100-115$.
Here,lower limit $(l) = 100$,frequency $(f) = 9$,cumulative frequency of the class preceding the median class $(cf) = 11$,and class width $(h) = 15$.
Using the median formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$= 100 + \left( \frac{16.5 - 11}{9} \right) \times 15$
$= 100 + \left( \frac{5.5}{9} \right) \times 15$
$= 100 + \frac{82.5}{9} = 100 + 9.166... \approx 109.17$.
Hence,the median bowling speed is $109.17 \, km/h$.

(D) In the given data,the highest frequency is $41,$ which corresponds to the modal class interval $10000-15000$.
Here,the lower limit of the modal class $l = 10000$,the frequency of the modal class $f_m = 41$,the frequency of the class preceding the modal class $f_1 = 26$,the frequency of the class succeeding the modal class $f_2 = 16$,and the class size $h = 5000$.
The formula for the mode is:
$\text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 10000 + \left( \frac{41 - 26}{2 \times 41 - 26 - 16} \right) \times 5000$
$= 10000 + \left( \frac{15}{82 - 42} \right) \times 5000$
$= 10000 + \left( \frac{15}{40} \right) \times 5000$
$= 10000 + 15 \times 125$
$= 10000 + 1875 = 11875$
Thus,the modal income is $Rs. 11875$.

(A) In the given data,the highest frequency is $26,$ which corresponds to the modal class interval $201-202$.
Here,the lower limit of the modal class $l = 201$,the frequency of the modal class $f_m = 26$,the frequency of the class preceding the modal class $f_1 = 12$,the frequency of the class succeeding the modal class $f_2 = 20$,and the class width $h = 1$.
The formula for the mode is:
$\text{Mode} = l + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$
Substituting the values:
$\text{Mode} = 201 + \left( \frac{26 - 12}{2 \times 26 - 12 - 20} \right) \times 1$
$= 201 + \left( \frac{14}{52 - 32} \right)$
$= 201 + \frac{14}{20} = 201 + 0.7 = 201.7 \, g$.
Hence,the modal weight is $201.7 \, g$.

(N/A) First,we convert the cumulative frequency distribution into a standard frequency distribution. The class interval starts from $20-30$ as the age is $20$ and above.

Class Interval	Frequency $(f_i)$	Class mark $(x_i)$	$u_i = \frac{x_i - 45}{10}$	$f_i u_i$
$20-30$	$100$	$25$	$-2$	$-200$
$30-40$	$120$	$35$	$-1$	$-120$
$40-50$	$130$	$45$	$0$	$0$
$50-60$	$400$	$55$	$1$	$400$
$60-70$	$200$	$65$	$2$	$400$
$70-80$	$50$	$75$	$3$	$150$
Total	$\sum f_i = 1000$	-	-	$\sum f_i u_i = 630$

Using the step-deviation method: $\text{Mean} (\bar{x}) = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right)$
Here,$a = 45$,$h = 10$,$\sum f_i u_i = 630$,and $\sum f_i = 1000$.
$\bar{x} = 45 + 10 \left( \frac{630}{1000} \right) = 45 + 6.3 = 51.3$.
Thus,the mean age is $51.3$ years.

(A) Given total frequency $\Sigma f_i = 100$.
Sum of frequencies: $17 + f_1 + 32 + f_2 + 19 = 68 + f_1 + f_2 = 100 \implies f_1 + f_2 = 32$ (Equation $1$).
Using the step-deviation method: $\bar{x} = A + \frac{\Sigma f_i u_i}{\Sigma f_i} \times c$,where $A = 65$ (assumed mean),$c = 20$ (class size).

Class	Frequency $(f_i)$	Midpoint $(x_i)$	$u_i = \frac{x_i - 65}{20}$	$f_i u_i$
$15-35$	$17$	$25$	$-2$	$-34$
$35-55$	$f_1$	$45$	$-1$	$-f_1$
$55-75$	$32$	$65$	$0$	$0$
$75-95$	$f_2$	$85$	$1$	$f_2$
$95-115$	$19$	$105$	$2$	$38$
Total	$100$	-	-	$f_2 - f_1 + 4$

Mean $\bar{x} = 65 + \frac{f_2 - f_1 + 4}{100} \times 20 = 65$.
$65 + \frac{f_2 - f_1 + 4}{5} = 65 \implies f_2 - f_1 + 4 = 0 \implies f_1 - f_2 = 4$ (Equation $2$).
Adding $(1)$ and $(2)$: $2f_1 = 36 \implies f_1 = 18$.
Substituting $f_1 = 18$ in $(1)$: $18 + f_2 = 32 \implies f_2 = 14$.
Thus,$f_1 = 18$ and $f_2 = 14$.

(D) To find the mean,we first calculate the class mark $(x_i)$ for each interval,where $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$.
$1$. For $21-25$,$x_1 = 23$,$f_1 = 18$,$f_1x_1 = 414$.
$2$. For $26-30$,$x_2 = 28$,$f_2 = 32$,$f_2x_2 = 896$.
$3$. For $31-35$,$x_3 = 33$,$f_3 = 30$,$f_3x_3 = 990$.
$4$. For $36-40$,$x_4 = 38$,$f_4 = 40$,$f_4x_4 = 1520$.
$5$. For $41-45$,$x_5 = 43$,$f_5 = 25$,$f_5x_5 = 1075$.
$6$. For $46-50$,$x_6 = 48$,$f_6 = 15$,$f_6x_6 = 720$.
$7$. For $51-55$,$x_7 = 53$,$f_7 = 40$,$f_7x_7 = 2120$.
Sum of frequencies $(\sum f_i)$ = $18 + 32 + 30 + 40 + 25 + 15 + 40 = 200$.
Sum of products $(\sum f_ix_i)$ = $414 + 896 + 990 + 1520 + 1075 + 720 + 2120 = 7735$.
Mean $(\bar{x})$ = $\frac{\sum f_ix_i}{\sum f_i} = \frac{7735}{200} = 38.675$.

(A) To calculate the mean,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$,where $x_i$ is the class mark.
$1$. Find the class marks $(x_i)$ for each interval:
- For $0-10$,$x_1 = (0+10)/2 = 5$
- For $10-20$,$x_2 = (10+20)/2 = 15$
- For $20-30$,$x_3 = (20+30)/2 = 25$
- For $30-40$,$x_4 = (30+40)/2 = 35$
- For $40-50$,$x_5 = (40+50)/2 = 45$
$2$. Calculate $f_i x_i$:
- $20 \times 5 = 100$
- $10 \times 15 = 150$
- $20 \times 25 = 500$
- $30 \times 35 = 1050$
- $20 \times 45 = 900$
$3$. Calculate $\sum f_i x_i = 100 + 150 + 500 + 1050 + 900 = 2700$.
$4$. Calculate $\sum f_i = 20 + 10 + 20 + 30 + 20 = 100$.
$5$. Mean $\bar{x} = 2700 / 100 = 27$.

(B) To find the mean,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$,where $x_i$ is the class mark.
$1$. Calculate class marks $(x_i)$:
$15, 25, 35, 45, 55, 65$.
$2$. Calculate $f_i x_i$:
$2 \times 15 = 30$
$10 \times 25 = 250$
$40 \times 35 = 1400$
$25 \times 45 = 1125$
$13 \times 55 = 715$
$10 \times 65 = 650$
$3$. Sum of frequencies $(\sum f_i)$: $2 + 10 + 40 + 25 + 13 + 10 = 100$.
$4$. Sum of products $(\sum f_i x_i)$: $30 + 250 + 1400 + 1125 + 715 + 650 = 4170$.
$5$. Mean $(\bar{x})$: $\frac{4170}{100} = 41.7$.