Mean of Grouped Data 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 $69.43 \%$.

(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.

(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 $u_{i} = \frac{x_{i} - a}{h}$.

(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.

(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 $d_i = x_i - a$.

(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 $Rs. 170.18$.

(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) 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 $(\bar{x})$: $\frac{4170}{100} = 41.7$.

(C) 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 $(x_i)$: $5, 15, 25, 35, 45, 55, 65, 75$.
$2$. Calculate $f_i x_i$:
$2 \times 5 = 10$
$4 \times 15 = 60$
$10 \times 25 = 250$
$20 \times 35 = 700$
$18 \times 45 = 810$
$20 \times 55 = 1100$
$16 \times 65 = 1040$
$10 \times 75 = 750$
$3$. Sum of frequencies $(\sum f_i)$ = $2 + 4 + 10 + 20 + 18 + 20 + 16 + 10 = 100$.
$4$. Sum of $f_i x_i$ $(\sum f_i x_i)$ = $10 + 60 + 250 + 700 + 810 + 1100 + 1040 + 750 = 4720$.
$5$. Mean $\bar{x} = \frac{4720}{100} = 47.2$.

(D) 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.
$1$. Calculate class marks $(x_i)$: $85, 95, 105, 115, 125, 135, 145, 155, 165$.
$2$. Calculate $f_i x_i$:
$6 \times 85 = 510$
$18 \times 95 = 1710$
$78 \times 105 = 8190$
$80 \times 115 = 9200$
$100 \times 125 = 12500$
$72 \times 135 = 9720$
$0 \times 145 = 0$
$40 \times 155 = 6200$
$6 \times 165 = 990$
$3$. Sum of frequencies $\sum f_i = 6 + 18 + 78 + 80 + 100 + 72 + 0 + 40 + 6 = 400$.
$4$. Sum of products $\sum f_i x_i = 510 + 1710 + 8190 + 9200 + 12500 + 9720 + 0 + 6200 + 990 = 49020$.
$5$. Mean $\bar{x} = \frac{49020}{400} = 122.55$.

(A) To calculate the mean,we first find the class mark $(x_i)$ for each class interval using the formula: $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$.
$1$. For $10-20$: $x_1 = \frac{10+20}{2} = 15$
$2$. For $20-30$: $x_2 = \frac{20+30}{2} = 25$
$3$. For $30-40$: $x_3 = \frac{30+40}{2} = 35$
$4$. For $40-50$: $x_4 = \frac{40+50}{2} = 45$
$5$. For $50-60$: $x_5 = \frac{50+60}{2} = 55$
Now,calculate $f_i x_i$ for each class:
- $12 \times 15 = 180$
- $16 \times 25 = 400$
- $8 \times 35 = 280$
- $6 \times 45 = 270$
- $8 \times 55 = 440$
Sum of frequencies $(\sum f_i)$: $12 + 16 + 8 + 6 + 8 = 50$.
Sum of products $(\sum f_i x_i)$: $180 + 400 + 280 + 270 + 440 = 1570$.
Mean $(\bar{x})$ = $\frac{\sum f_i x_i}{\sum f_i} = \frac{1570}{50} = 31.4$.

(B) To find the mean,we first calculate the class mark $(x_i)$ for each interval:
$11-13: x_1 = 12$
$13-15: x_2 = 14$
$15-17: x_3 = 16$
$17-19: x_4 = 18$
$19-21: x_5 = 20$
$21-23: x_6 = 22$
$23-25: x_7 = 24$
Now,calculate $f_i x_i$:
$3 \times 12 = 36$
$6 \times 14 = 84$
$9 \times 16 = 144$
$13 \times 18 = 234$
$f \times 20 = 20f$
$5 \times 22 = 110$
$4 \times 24 = 96$
Sum of frequencies $\sum f_i = 3+6+9+13+f+5+4 = 40+f$
Sum of products $\sum f_i x_i = 36+84+144+234+20f+110+96 = 704+20f$
Given mean $\bar{x} = 18$,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$:
$18 = \frac{704+20f}{40+f}$
$18(40+f) = 704+20f$
$720+18f = 704+20f$
$720-704 = 20f-18f$
$16 = 2f$
$f = 8$

(C) To find the mean,we first calculate the class mark $(x_i)$ for each interval:
$10-20: x_1 = 15$
$20-30: x_2 = 25$
$30-40: x_3 = 35$
$40-50: x_4 = 45$
$50-60: x_5 = 55$
$60-70: x_6 = 65$
$70-80: x_7 = 75$
The sum of frequencies $\sum f_i = 5 + 3 + 4 + f + 2 + 6 + 13 = 33 + f$.
The sum of products $\sum f_i x_i = (5 \times 15) + (3 \times 25) + (4 \times 35) + (f \times 45) + (2 \times 55) + (6 \times 65) + (13 \times 75)$
$= 75 + 75 + 140 + 45f + 110 + 390 + 975 = 1765 + 45f$.
Given mean $\bar{x} = 52$,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$:
$52 = \frac{1765 + 45f}{33 + f}$
$52(33 + f) = 1765 + 45f$
$1716 + 52f = 1765 + 45f$
$52f - 45f = 1765 - 1716$
$7f = 49$
$f = 7$.

(A) Given total frequency $\sum f_i = 120$. Thus,$17 + f_1 + 32 + f_2 + 19 = 120 \implies f_1 + f_2 = 52$ (Equation $1$).
The class marks $(x_i)$ are $20, 40, 60, 80, 100$. The sum $\sum f_i x_i = (17 \times 20) + (f_1 \times 40) + (32 \times 60) + (f_2 \times 80) + (19 \times 100) = 340 + 40f_1 + 1920 + 80f_2 + 1900 = 4160 + 40f_1 + 80f_2$.
Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 60 \implies \frac{4160 + 40f_1 + 80f_2}{120} = 60$.
$4160 + 40f_1 + 80f_2 = 7200 \implies 40f_1 + 80f_2 = 3040 \implies f_1 + 2f_2 = 76$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(f_1 + 2f_2) - (f_1 + f_2) = 76 - 52 \implies f_2 = 24$.
Substituting $f_2 = 24$ in Equation $1$: $f_1 + 24 = 52 \implies f_1 = 28$. Thus,$f_1 = 28$ and $f_2 = 24$.

(B) To find the mean of a grouped frequency distribution,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
First,calculate the class marks $(x_i)$ for each interval:
$0-20: x_1 = 10$
$20-40: x_2 = 30$
$40-60: x_3 = 50$
$60-80: x_4 = 70$
$80-100: x_5 = 90$
$100-120: x_6 = 110$
Now,calculate $\sum f_i x_i = (10 \times 10) + (f \times 30) + (7 \times 50) + (6 \times 70) + (5 \times 90) + (4 \times 110) = 100 + 30f + 350 + 420 + 450 + 440 = 1760 + 30f$.
Calculate $\sum f_i = 10 + f + 7 + 6 + 5 + 4 = 32 + f$.
Given the mean $\bar{x} = 46$,we have $46 = \frac{1760 + 30f}{32 + f}$.
$46(32 + f) = 1760 + 30f$
$1472 + 46f = 1760 + 30f$
$46f - 30f = 1760 - 1472$
$16f = 288$
$f = \frac{288}{16} = 18$.

(A) To find the mean,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
First,calculate the class marks $(x_i)$ for each interval: $2, 6, 10, 14, 18, 22, 26, 30, 34$.
Next,calculate $f_i x_i$ for each class:
$6 \times 2 = 12$
$8 \times 6 = 48$
$17 \times 10 = 170$
$23 \times 14 = 322$
$16 \times 18 = 288$
$15 \times 22 = 330$
$f \times 26 = 26f$
$4 \times 30 = 120$
$3 \times 34 = 102$
Sum of frequencies $\sum f_i = 6+8+17+23+16+15+f+4+3 = 92+f$.
Sum of $f_i x_i = 12+48+170+322+288+330+26f+120+102 = 1392+26f$.
Given mean $\bar{x} = 16$,so $16 = \frac{1392+26f}{92+f}$.
$16(92+f) = 1392+26f$
$1472 + 16f = 1392 + 26f$
$1472 - 1392 = 26f - 16f$
$80 = 10f$
$f = 8$.

(D) In the assumed mean method for calculating the mean,the formula is given by $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$.
Here,$A$ is the assumed mean and $x_{i}$ is the class mark (midpoint) of the $i^{th}$ class interval.
The deviation $d_{i}$ is defined as the difference between the class mark and the assumed mean.
Therefore,$d_{i} = x_{i} - A$.

(A) The formula provided is for the step-deviation method to calculate the mean of grouped data.
In this method,the deviation $d_{i}$ is defined as $d_{i} = x_{i} - A$,where $x_{i}$ is the class mark and $A$ is the assumed mean.
The variable $u_{i}$ is defined as the ratio of the deviation to the class width $c$,which is given by $u_{i} = \frac{x_{i} - A}{c}$.
Therefore,the correct expression for $u_{i}$ is $\frac{x_{i} - A}{c}$.

(A) Let the ages of Ina,Mina,and Dika be $I, M,$ and $D$ respectively.
The average age is given by $\frac{I + M + D}{3} = x$.
Therefore,the sum of their current ages is $I + M + D = 3x$.
After $y$ years,each person's age will increase by $y$ years.
The new sum of their ages will be $(I + y) + (M + y) + (D + y) = (I + M + D) + 3y$.
Substituting the sum of current ages: $3x + 3y$.
The new average age will be $\frac{3x + 3y}{3} = x + y$.

(B) The mean (average) of a set of $n$ observations $x_1, x_2, ..., x_n$ is defined as the sum of all observations divided by the total number of observations.
Mathematically,this is represented as $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$,where $\sum x_i$ is the sum of all values and $n$ is the total count of observations.

(A) The mean $(\bar{x})$ of grouped data using the direct method is calculated by the formula: $\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$,where $f_{i}$ is the frequency of the $i^{th}$ class interval and $x_{i}$ is the class mark (midpoint) of the $i^{th}$ class interval. The sum $\sum f_{i}$ is often denoted by $N$ or $n$,representing the total number of observations.

(B) In the given formula for the mean of grouped data,$\bar{x} = \frac{\Sigma f_{i} x_{i}}{n}$:
$\bar{x}$ represents the mean.
$f_{i}$ represents the frequency of the $i^{th}$ class.
$x_{i}$ represents the class mark (midvalue) of the $i^{th}$ class.
$n$ represents the total frequency,which is equal to $\Sigma f_{i}$.
Therefore,$f_{i}$ represents the frequency of the $i^{th}$ class.