Mix Examples - Statistics Questions in English

(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 $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$:
$52 = \frac{1765 + 45f}{33 + f}$
$52(33 + f) = 1765 + 45f$
$1716 + 52f = 1765 + 45f$
$52f - 45f = 1765 - 1716$
$7f = 49$
$f = 7$.

(D) Given total frequency $N = 60$.
Sum of frequencies: $6 + f_1 + 20 + 15 + f_2 + 4 = 60$
$45 + f_1 + f_2 = 60 \implies f_1 + f_2 = 15$ ---(Equation $1$)
Calculate midpoint $(x_i)$ and $f_i x_i$:
Class $0-10$: $x_i = 5, f_i x_i = 30$
Class $10-20$: $x_i = 15, f_i x_i = 15f_1$
Class $20-30$: $x_i = 25, f_i x_i = 500$
Class $30-40$: $x_i = 35, f_i x_i = 525$
Class $40-50$: $x_i = 45, f_i x_i = 45f_2$
Class $50-60$: $x_i = 55, f_i x_i = 220$
Sum $\sum f_i x_i = 30 + 15f_1 + 500 + 525 + 45f_2 + 220 = 1275 + 15f_1 + 45f_2$
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = 28.5$
$\frac{1275 + 15f_1 + 45f_2}{60} = 28.5$
$1275 + 15f_1 + 45f_2 = 1710$
$15f_1 + 45f_2 = 435$
Divide by $15$: $f_1 + 3f_2 = 29$ ---(Equation $2$)
Subtract Equation $1$ from Equation $2$:
$(f_1 + 3f_2) - (f_1 + f_2) = 29 - 15$
$2f_2 = 14 \implies f_2 = 7$
Substitute $f_2 = 7$ in Equation $1$: $f_1 + 7 = 15 \implies f_1 = 8$.
Thus,$f_1 = 8$ and $f_2 = 7$.

(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) The modal class is the class with the highest frequency. Here,the highest frequency is $37$,which corresponds to the class interval $200-300$.
Thus,the modal class is $200-300$.
We use the formula for mode: $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ (lower limit of modal class) $= 200$
$h$ (class size) $= 100$
$f_1$ (frequency of modal class) $= 37$
$f_0$ (frequency of the class preceding the modal class) $= 21$
$f_2$ (frequency of the class succeeding the modal class) $= 13$
Substituting these values into the formula:
$Z = 200 + \left( \frac{37 - 21}{2(37) - 21 - 13} \right) \times 100$
$Z = 200 + \left( \frac{16}{74 - 34} \right) \times 100$
$Z = 200 + \left( \frac{16}{40} \right) \times 100$
$Z = 200 + 0.4 \times 100$
$Z = 200 + 40 = 240$
Therefore,the mode of the given frequency distribution is $240$.

(A) Here,the mode $33 \frac{1}{3}$ lies in the class $30-40$.
Therefore,$30-40$ is the modal class.
Now,$c$ (class length) $= 10$,$l$ (lower limit of the modal class) $= 30$,$f_1$ (frequency of the modal class) $= 28$,$f_0$ (frequency of the class preceding the modal class) $= x$,and $f_2$ (frequency of the class succeeding the modal class) $= y$.
Substituting the values in the mode formula $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times c$:
$33 \frac{1}{3} = 30 + \left( \frac{28 - x}{2(28) - x - y} \right) \times 10$
$3 \frac{1}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{10}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{1}{3} = \frac{28 - x}{56 - x - y}$
$56 - x - y = 84 - 3x$
$2x - y = 28$ ... $(1)$
Moreover,the total frequency is $100$.
$7 + 12 + x + 28 + y + 9 = 100$
$56 + x + y = 100$
$x + y = 44$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$3x = 72 \implies x = 24$
Substituting $x = 24$ in equation $(2)$:
$24 + y = 44 \implies y = 20$
Thus,the missing frequencies are $x = 24$ and $y = 20$.

(D) To find the mode of a grouped frequency distribution,we use the formula:
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class (the class with the highest frequency). Here,the highest frequency is $55$,which corresponds to the class interval $60-70$.
$2$. From the modal class:
Lower limit $(l)$ = $60$
Frequency of the modal class $(f_1)$ = $55$
Frequency of the class preceding the modal class $(f_0)$ = $43$
Frequency of the class succeeding the modal class $(f_2)$ = $37$
Class size $(h)$ = $70 - 60 = 10$
$3$. Substitute the values into the formula:
$Mode = 60 + \left( \frac{55 - 43}{2(55) - 43 - 37} \right) \times 10$
$Mode = 60 + \left( \frac{12}{110 - 80} \right) \times 10$
$Mode = 60 + \left( \frac{12}{30} \right) \times 10$
$Mode = 60 + \left( \frac{12}{3} \right)$
$Mode = 60 + 4 = 64$
Therefore,the mode of the given frequency distribution is $64$.

(A) To find the mode of a grouped frequency distribution,we use the formula:
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class: The class with the highest frequency is $40-60$ (frequency $26$).
$2$. Here,the lower limit of the modal class $l = 40$,the size of the class interval $h = 20$.
$3$. The frequency of the modal class $f_1 = 26$.
$4$. The frequency of the class preceding the modal class $f_0 = 9$.
$5$. The frequency of the class succeeding the modal class $f_2 = 23$.
$6$. Substituting these values into the formula:
$Mode = 40 + \left( \frac{26 - 9}{2(26) - 9 - 23} \right) \times 20$
$Mode = 40 + \left( \frac{17}{52 - 32} \right) \times 20$
$Mode = 40 + \left( \frac{17}{20} \right) \times 20$
$Mode = 40 + 17 = 57$
Thus,the mode is $57$.

(B) The formula for the mode of a grouped frequency distribution is given by:
$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$1$. Identify the modal class: The class with the highest frequency is $40-50$,so the modal class is $40-50$.
$2$. Identify the parameters:
- Lower limit of the modal class $(l)$ = $40$
- Frequency of the modal class $(f_1)$ = $42$
- Frequency of the class preceding the modal class $(f_0)$ = $24$
- Frequency of the class succeeding the modal class $(f_2)$ = $30$
- Class size $(h)$ = $10$
$3$. Substitute the values into the formula:
$Mode = 40 + \left( \frac{42 - 24}{2(42) - 24 - 30} \right) \times 10$
$Mode = 40 + \left( \frac{18}{84 - 54} \right) \times 10$
$Mode = 40 + \left( \frac{18}{30} \right) \times 10$
$Mode = 40 + \left( \frac{18}{3} \right) = 40 + 6 = 46$
Thus,the mode is $46$.

(C) The modal class is the class with the highest frequency. Here,the highest frequency is $66$,which corresponds to the class interval $60-75$.
Formula for mode: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Here:
$l$ (lower limit of modal class) $= 60$
$h$ (class size) $= 15$
$f_1$ (frequency of modal class) $= 66$
$f_0$ (frequency of the class preceding the modal class) $= 45$
$f_2$ (frequency of the class succeeding the modal class) $= 42$
Substituting the values:
$\text{Mode} = 60 + \left( \frac{66 - 45}{2(66) - 45 - 42} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{132 - 87} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{45} \right) \times 15$
$\text{Mode} = 60 + \left( \frac{21}{3} \right) = 60 + 7 = 67$

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

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$65-85$	$4$	$4$
$85-105$	$5$	$9$
$105-125$	$13$	$22$
$125-145$	$20$	$42$
$145-165$	$14$	$56$
$165-185$	$8$	$64$
$185-205$	$4$	$68$

Here,total frequency $n = 68$.
$\therefore \frac{n}{2} = \frac{68}{2} = 34$.
The cumulative frequency just greater than $34$ is $42$,which corresponds to the class interval $125-145$.
Thus,the median class is $125-145$.
Here,$l = 125$,$cf = 22$,$f = 20$,and class size $h = 20$.
Using the median formula: $M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$M = 125 + \left( \frac{34 - 22}{20} \right) \times 20$
$M = 125 + 12 = 137$.
Therefore,the median of the given frequency distribution is $137$.

(X=8, Y=7)

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

Given total frequency $n = 60$.
Therefore,$45 + x + y = 60 \implies x + y = 15$ (Equation $1$).
Since $n = 60$,$\frac{n}{2} = 30$.
The median $28.5$ lies in the class $20-30$. Thus,the median class is $20-30$.
Here,$l = 20$,$cf = 5 + x$,$f = 20$,and class size $h = 10$.
Using the median formula: $Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10$
$8.5 = \frac{25 - x}{2}$
$17 = 25 - x \implies x = 8$.
Substituting $x = 8$ in Equation $1$: $8 + y = 15 \implies y = 7$.
The missing frequencies are $x = 8$ and $y = 7$.

(B)

Class	Frequency $(f)$	Cumulative frequency $(cf)$
$5-10$	$5$	$5$
$10-15$	$6$	$11$
$15-20$	$15$	$26$
$20-25$	$10$	$36$
$25-30$	$5$	$41$
$30-35$	$4$	$45$
$35-40$	$2$	$47$
$40-45$	$2$	$49$

Here,the total frequency $n = 49$.
$\therefore \frac{n}{2} = \frac{49}{2} = 24.5$.
The cumulative frequency just greater than $24.5$ is $26$,which corresponds to the class interval $15-20$.
Therefore,the median class is $15-20$.
Here,$l = 15$ (lower limit of median class),$cf = 11$ (cumulative frequency of the class preceding the median class),$f = 15$ (frequency of the median class),and $h = 5$ (class size).
Using the median formula: $M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$M = 15 + \left( \frac{24.5 - 11}{15} \right) \times 5$
$M = 15 + \left( \frac{13.5}{15} \right) \times 5$
$M = 15 + \frac{13.5}{3} = 15 + 4.5 = 19.5$.
Thus,the median of the given frequency distribution is $19.5$.

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

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$30-35$	$1$	$1$
$35-40$	$2$	$3$
$40-45$	$5$	$8$
$45-50$	$10$	$18$
$50-55$	$17$	$35$
$55-60$	$15$	$50$
$60-65$	$9$	$59$
$65-70$	$5$	$64$
$70-75$	$3$	$67$

Total frequency $N = 67$. So,$N/2 = 67/2 = 33.5$.
Looking at the $cf$ column,the median class is $50-55$ because $33.5$ lies in this interval.
Here,lower limit $l = 50$,class size $h = 5$,frequency of median class $f = 17$,and cumulative frequency of the class preceding the median class $cf = 18$.
Using the median formula: $\text{Median} = l + \left( \frac{N/2 - cf}{f} \right) \times h$
$\text{Median} = 50 + \left( \frac{33.5 - 18}{17} \right) \times 5$
$\text{Median} = 50 + \left( \frac{15.5}{17} \right) \times 5$
$\text{Median} = 50 + \frac{77.5}{17} \approx 50 + 4.5588 \approx 54.56$.

(D) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$60-70$	$5$	$5$
$70-80$	$15$	$20$
$80-90$	$20$	$40$
$90-100$	$30$	$70$
$100-110$	$20$	$90$
$110-120$	$8$	$98$

Total frequency $N = 98$. So,$N/2 = 98/2 = 49$.
The cumulative frequency just greater than $49$ is $70$,which corresponds to the class interval $90-100$. Thus,the median class is $90-100$.
Here,$l = 90$,$f = 30$,$cf = 40$,and $h = 10$.
Using the formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$\text{Median} = 90 + \left( \frac{49 - 40}{30} \right) \times 10$
$\text{Median} = 90 + \left( \frac{9}{30} \right) \times 10 = 90 + 3 = 93$.

(A) To find the median,we first calculate the cumulative frequency $(cf)$:
| Class | Frequency $(f)$ | Cumulative Frequency $(cf)$ |
| :--- | :--- | :--- |
| $110-120$ | $6$ | $6$ |
| $120-130$ | $25$ | $31$ |
| $130-140$ | $48$ | $79$ |
| $140-150$ | $72$ | $151$ |
| $150-160$ | $116$ | $267$ |
| $160-170$ | $60$ | $327$ |
| $170-180$ | $38$ | $365$ |
| $180-190$ | $22$ | $387$ |
| $190-200$ | $3$ | $390$ |
Total frequency $N = 390$. So,$N/2 = 390/2 = 195$.
The cumulative frequency just greater than $195$ is $267$,which corresponds to the class interval $150-160$.
Thus,the median class is $150-160$.
Here,lower limit $l = 150$,frequency of median class $f = 116$,cumulative frequency of the class preceding the median class $cf = 151$,and class size $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{N/2 - cf}{f} \right) \times h$
$\text{Median} = 150 + \left( \frac{195 - 151}{116} \right) \times 10$
$\text{Median} = 150 + \left( \frac{44}{116} \right) \times 10 = 150 + \frac{440}{116} \approx 150 + 3.79 = 153.79$.

(B) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-20$	$5$	$5$
$20-40$	$15$	$20$
$40-60$	$20$	$40$
$60-80$	$18$	$58$
$80-100$	$2$	$60$

Total frequency $N = 60$. So,$N/2 = 60/2 = 30$.
The cumulative frequency just greater than $30$ is $40$,which corresponds to the median class $40-60$.
Here,lower limit $l = 40$,class size $h = 20$,frequency of median class $f = 20$,and cumulative frequency of the class preceding the median class $cf = 20$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$\text{Median} = 40 + \left( \frac{30 - 20}{20} \right) \times 20$
$\text{Median} = 40 + \left( \frac{10}{20} \right) \times 20 = 40 + 10 = 50$.

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

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-10$	$2$	$2$
$10-20$	$6$	$8$
$20-30$	$8$	$16$
$30-40$	$16$	$32$
$40-50$	$20$	$52$
$50-60$	$18$	$70$
$60-70$	$16$	$86$
$70-80$	$14$	$100$

Total frequency $N = 100$. So,$N/2 = 50$.
The cumulative frequency just greater than $50$ is $52$,which corresponds to the class interval $40-50$.
Thus,the median class is $40-50$.
Here,$l = 40$,$f = 20$,$cf = 32$,and $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{N/2 - cf}{f} \right) \times h$
$\text{Median} = 40 + \left( \frac{50 - 32}{20} \right) \times 10$
$\text{Median} = 40 + \left( \frac{18}{20} \right) \times 10 = 40 + 9 = 49$.

(D) To find the median,we first calculate the cumulative frequency $(cf)$:

Class	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-5$	$5$	$5$
$5-10$	$15$	$20$
$10-15$	$15$	$35$
$15-20$	$8$	$43$
$20-25$	$4$	$47$
$25-30$	$2$	$49$
$30-35$	$1$	$50$

Total frequency $N = 50$. So,$N/2 = 25$.
The cumulative frequency just greater than $25$ is $35$,which corresponds to the class interval $10-15$.
Thus,the median class is $10-15$.
Here,$l = 10$,$f = 15$,$cf$ (of preceding class) $= 20$,and $h = 5$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$\text{Median} = 10 + \left( \frac{25 - 20}{15} \right) \times 5$
$\text{Median} = 10 + \left( \frac{5}{15} \right) \times 5 = 10 + \frac{25}{15} = 10 + 1.666... = 11.67$.

(A) Given,total frequency $N = 100$.
Sum of frequencies: $2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100$
$76 + x + y = 100 \implies x + y = 24$ --- $(1)$
Since the median is $525$,the median class is $500-600$.
Here,$l = 500$,$f = 20$,$cf = (2 + 5 + x + 12 + 17) = 36 + x$,$h = 100$,and $N/2 = 50$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x \implies x = 9$
Substituting $x = 9$ in $(1)$: $9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.

(N/A) Given total frequency $N = 230$.
Sum of frequencies: $12 + 30 + x + 65 + y + 25 + 18 = 230$
$150 + x + y = 230 \implies x + y = 80$ ---(Equation $1$)
Since the median is $46$,the median class is $40-50$.
Here,$l = 40$,$h = 10$,$f = 65$,$N = 230$,and cumulative frequency of the class preceding the median class $cf = 12 + 30 + x = 42 + x$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$46 = 40 + \left( \frac{115 - (42 + x)}{65} \right) \times 10$
$6 = \left( \frac{73 - x}{65} \right) \times 10$
$6 \times 6.5 = 73 - x$
$39 = 73 - x \implies x = 34$.
Substituting $x = 34$ in Equation $1$: $34 + y = 80 \implies y = 46$.
Thus,$x = 34$ and $y = 46$.

(N/A) $1$. Mean: The class marks $(x_i)$ are $15, 45, 75, 105, 135, 165$. The sum of frequencies $(sum f_i)$ is $80$. The sum of products $(sum f_i x_i)$ is $(8 \times 15) + (15 \times 45) + (16 \times 75) + (20 \times 105) + (12 \times 135) + (9 \times 165) = 120 + 675 + 1200 + 2100 + 1620 + 1485 = 7200$. Mean $= \frac{\sum f_i x_i}{\sum f_i} = \frac{7200}{80} = 90$.
$2$. Median: $N/2 = 40$. The cumulative frequencies are $8, 23, 39, 59, 71, 80$. The median class is $90-120$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 90 + \left( \frac{40 - 39}{20} \right) \times 30 = 90 + (1/20) \times 30 = 90 + 1.5 = 91.5$.
$3$. Mode: The modal class is $90-120$ (highest frequency $20$). Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 90 + \left( \frac{20 - 16}{2(20) - 16 - 12} \right) \times 30 = 90 + \left( \frac{4}{40 - 28} \right) \times 30 = 90 + \left( \frac{4}{12} \right) \times 30 = 90 + 10 = 100$.

(N/A) $1$. Mean: $\sum f_i = 100$. Midpoints $(x_i)$: $5, 15, 25, 35, 45, 55, 65$. $\sum f_i x_i = (5 \times 5) + (12 \times 15) + (18 \times 25) + (40 \times 35) + (15 \times 45) + (7 \times 55) + (3 \times 65) = 25 + 180 + 450 + 1400 + 675 + 385 + 195 = 3310$. Mean $= \frac{\sum f_i x_i}{\sum f_i} = \frac{3310}{100} = 33.1$.
$2$. Median: $N=100, N/2 = 50$. Cumulative frequencies: $5, 17, 35, 75, 90, 97, 100$. Median class is $30-40$. $l=30, cf=35, f=40, h=10$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 30 + \left( \frac{50 - 35}{40} \right) \times 10 = 30 + \left( \frac{15}{4} \right) = 30 + 3.75 = 33.75$.
$3$. Mode: Modal class is $30-40$ (highest frequency $40$). $l=30, f_1=40, f_0=18, f_2=15, h=10$. Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 30 + \left( \frac{40 - 18}{2(40) - 18 - 15} \right) \times 10 = 30 + \left( \frac{22}{80 - 33} \right) \times 10 = 30 + \left( \frac{220}{47} \right) \approx 30 + 4.68 = 34.68$.

(N/A) The total number of students is $N = 400$.
Therefore,$26 + 26 + x + 110 + 84 + y + 36 + 32 = 400$.
$314 + x + y = 400 \implies x + y = 86$ ---$(i)$
The mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 41.2$.
The class marks $(x_i)$ are: $5, 15, 25, 35, 45, 55, 65, 75$.
The sum of $f_i x_i$ is: $(26 \times 5) + (26 \times 15) + (x \times 25) + (110 \times 35) + (84 \times 45) + (y \times 55) + (36 \times 65) + (32 \times 75)$.
$= 130 + 390 + 25x + 3850 + 3780 + 55y + 2340 + 2400 = 12890 + 25x + 55y$.
Thus,$\frac{12890 + 25x + 55y}{400} = 41.2$.
$12890 + 25x + 55y = 16480 \implies 25x + 55y = 3590$.
Dividing by $5$,we get $5x + 11y = 718$ ---(ii).
From equation $(i)$,$x = 86 - y$. Substituting this into $(ii)$:
$5(86 - y) + 11y = 718 \implies 430 - 5y + 11y = 718$.
$6y = 288 \implies y = 48$.
Therefore,$x = 86 - 48 = 38$.
Answer: $x = 38, y = 48$.

(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) Given total frequency $N = 60$.
Sum of frequencies: $6 + x + 17 + y + 8 = 60 \implies x + y + 31 = 60 \implies x + y = 29$ (Equation $1$).
The class marks $(f_i)$ are $5, 15, 25, 35, 45$.
Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 26.5$.
$\sum f_i x_i = (6 \times 5) + (x \times 15) + (17 \times 25) + (y \times 35) + (8 \times 45) = 30 + 15x + 425 + 35y + 360 = 815 + 15x + 35y$.
$\frac{815 + 15x + 35y}{60} = 26.5 \implies 815 + 15x + 35y = 1590 \implies 15x + 35y = 775$.
Dividing by $5$: $3x + 7y = 155$ (Equation $2$).
From Equation $1$,$x = 29 - y$. Substituting into Equation $2$: $3(29 - y) + 7y = 155 \implies 87 - 3y + 7y = 155 \implies 4y = 68 \implies y = 17$.
Then $x = 29 - 17 = 12$.
Thus,$x = 12$ and $y = 17$.

(A) Given total frequency $N = 20$.
Sum of frequencies: $3 + x + 7 + 7 + y = 20 \implies x + y + 17 = 20 \implies x + y = 3$ (Equation $1$).
The class marks $(x_i)$ are $35, 45, 55, 65, 75$.
The sum of $f_i x_i = (3 \times 35) + (x \times 45) + (7 \times 55) + (7 \times 65) + (y \times 75) = 105 + 45x + 385 + 455 + 75y = 945 + 45x + 75y$.
Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 55.5$.
$\frac{945 + 45x + 75y}{20} = 55.5 \implies 945 + 45x + 75y = 1110 \implies 45x + 75y = 165$.
Dividing by $15$: $3x + 5y = 11$ (Equation $2$).
From Equation $1$,$x = 3 - y$. Substituting into Equation $2$: $3(3 - y) + 5y = 11 \implies 9 - 3y + 5y = 11 \implies 2y = 2 \implies y = 1$.
Then $x = 3 - 1 = 2$.
Thus,$x = 2$ and $y = 1$.

(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 $16 = \frac{1392+26f}{92+f}$.
$16(92+f) = 1392+26f$
$1472 + 16f = 1392 + 26f$
$1472 - 1392 = 26f - 16f$
$80 = 10f$
$f = 8$.

(N/A) $1$. Mean: $\sum f_i = 50$. Mid-points $(x_i)$: $5, 15, 25, 35, 45, 55, 65$. $\sum f_i x_i = (4 \times 5) + (6 \times 15) + (8 \times 25) + (12 \times 35) + (10 \times 45) + (5 \times 55) + (5 \times 65) = 20 + 90 + 200 + 420 + 450 + 275 + 325 = 1780$. Mean $\bar{x} = \frac{1780}{50} = 35.6$.
$2$. Median: $N/2 = 25$. Cumulative frequencies: $4, 10, 18, 30, 40, 45, 50$. Median class is $30-40$. $l=30, f=12, cf=18, h=10$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 30 + \left( \frac{25 - 18}{12} \right) \times 10 = 30 + \frac{70}{12} = 30 + 5.83 = 35.83$.
$3$. Mode: Modal class is $30-40$ (highest frequency $12$). $l=30, f_1=12, f_0=8, f_2=10, h=10$. Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 30 + \left( \frac{12 - 8}{24 - 8 - 10} \right) \times 10 = 30 + \left( \frac{4}{6} \right) \times 10 = 30 + 6.67 = 36.67$.

(N/A) $1$. Mean Calculation:
Class marks $(x_i)$: $5, 15, 25, 35, 45$.
Sum of frequencies $(\sum f_i)$ $= 5 + 27 + 58 + 20 + 10 = 120$.
Sum of products $(\sum f_i x_i)$ $= (5 \times 5) + (27 \times 15) + (58 \times 25) + (20 \times 35) + (10 \times 45) = 25 + 405 + 1450 + 700 + 450 = 3030$.
Mean $(\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{3030}{120} = 25.25$.
$2$. Median Calculation:
Cumulative frequencies $(cf)$: $5, 32, 90, 110, 120$.
$N/2 = 120/2 = 60$. The median class is $20-30$.
$l = 20, cf = 32, f = 58, h = 10$.
Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 20 + \left( \frac{60 - 32}{58} \right) \times 10 = 20 + \left( \frac{280}{58} \right) \approx 20 + 4.83 = 24.83$.
$3$. Mode Calculation:
Modal class is $20-30$ (highest frequency $58$).
$l = 20, f_1 = 58, f_0 = 27, f_2 = 20, h = 10$.
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 20 + \left( \frac{58 - 27}{116 - 27 - 20} \right) \times 10 = 20 + \left( \frac{31}{69} \right) \times 10 \approx 20 + 4.49 = 24.49$.

(A) $1$. Mean: $\sum f_i = 100$. Mid-points $(x_i)$: $25, 75, 125, 175, 225, 275, 325$. $\sum f_i x_i = 250 + 1125 + 3750 + 3500 + 3375 + 2200 + 650 = 14850$. Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{14850}{100} = 148.5$.
$2$. Median: $N/2 = 50$. Cumulative frequencies: $10, 25, 55, 75, 90, 98, 100$. Median class is $100-150$. $l=100, cf=25, f=30, h=50$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 100 + \left( \frac{50-25}{30} \right) \times 50 = 100 + \frac{25 \times 50}{30} = 100 + 41.67 = 141.67$.
$3$. Mode: Modal class is $100-150$ $(f_1=30, f_0=15, f_2=20)$. Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 100 + \left( \frac{30-15}{60-15-20} \right) \times 50 = 100 + \left( \frac{15}{25} \right) \times 50 = 100 + 30 = 130$.

(N/A) $1$. Mean: The class marks $(x_i)$ are $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$. The sum of frequencies $(sum f_i)$ is $100$. The sum of products $(sum f_i x_i)$ is $(11 \times 7.5) + (20 \times 12.5) + (35 \times 17.5) + (20 \times 22.5) + (8 \times 27.5) + (6 \times 32.5) = 82.5 + 250 + 612.5 + 450 + 220 + 195 = 1810$. Mean $= \frac{\sum f_i x_i}{\sum f_i} = \frac{1810}{100} = 18.1$.
$2$. Median: $N/2 = 50$. The cumulative frequencies are $11, 31, 66, 86, 94, 100$. The median class is $15-20$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 15 + \left( \frac{50 - 31}{35} \right) \times 5 = 15 + \left( \frac{19}{35} \right) \times 5 = 15 + \frac{19}{7} \approx 15 + 2.71 = 17.71$.
$3$. Mode: The modal class is $15-20$ (highest frequency $35$). Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 15 + \left( \frac{35 - 20}{2(35) - 20 - 20} \right) \times 5 = 15 + \left( \frac{15}{70 - 40} \right) \times 5 = 15 + \left( \frac{15}{30} \right) \times 5 = 15 + 2.5 = 17.5$.

(N/A) $1$. Mean: The total frequency $N = 50$. The midpoints $(x_i)$ are $35, 45, 55, 65, 75, 85, 95$. The sum $\sum f_i x_i = (5 \times 35) + (7 \times 45) + (12 \times 55) + (10 \times 65) + (8 \times 75) + (6 \times 85) + (2 \times 95) = 175 + 315 + 660 + 650 + 600 + 510 + 190 = 3100$. Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{3100}{50} = 62$.
$2$. Median: $N/2 = 25$. The cumulative frequencies are $5, 12, 24, 34, 42, 48, 50$. The median class is $60-70$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 60 + \left( \frac{25 - 24}{10} \right) \times 10 = 60 + 1 = 61$.
$3$. Mode: The modal class is $50-60$ (highest frequency $12$). Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 50 + \left( \frac{12 - 7}{2(12) - 7 - 10} \right) \times 10 = 50 + \left( \frac{5}{24 - 17} \right) \times 10 = 50 + \frac{50}{7} \approx 50 + 7.14 = 57.14$.

(C) The mode is given as $37$. Since $37$ lies in the class interval $30-40$,the modal class is $30-40$.
The formula for mode is: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Here,$l = 30$ (lower limit of modal class),$h = 10$ (class size),$f_1 = 18$ (frequency of modal class),$f_0 = x$ (frequency of preceding class),and $f_2 = 15$ (frequency of succeeding class).
Substituting the values: $37 = 30 + \left( \frac{18 - x}{2(18) - x - 15} \right) \times 10$
$37 - 30 = \left( \frac{18 - x}{36 - 15 - x} \right) \times 10$
$7 = \left( \frac{18 - x}{21 - x} \right) \times 10$
$7(21 - x) = 10(18 - x)$
$147 - 7x = 180 - 10x$
$10x - 7x = 180 - 147$
$3x = 33$
$x = 11$.

(A) Given total frequency $N = 150$.
Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$
$111 + x + y = 150 \implies x + y = 39$ --- $(1)$
Since the mode is $46$,the modal class is $40-50$.
Here,$l = 40, f_1 = 42, f_0 = x, f_2 = 30, h = 10$.
Mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$46 = 40 + \left( \frac{42 - x}{2(42) - x - 30} \right) \times 10$
$6 = \left( \frac{42 - x}{84 - 30 - x} \right) \times 10$
$6 = \frac{10(42 - x)}{54 - x} \implies 3(54 - x) = 5(42 - x)$
$162 - 3x = 210 - 5x$
$2x = 48 \implies x = 24$
Substituting $x = 24$ in $(1)$: $24 + y = 39 \implies y = 15$.
Thus,$x = 24$ and $y = 15$.

(X=43, Y=18) Given total frequency $N = 200.$
Sum of frequencies: $8 + 12 + 27 + x + 55 + 37 + y = 200 \implies 139 + x + y = 200 \implies x + y = 61$ (Equation $1$).
Since the mode is $64,$ the modal class is $60-70.$
Formula for mode: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h.$
Here,$l = 60, f_1 = 55, f_0 = x, f_2 = 37, h = 10.$
$64 = 60 + \left( \frac{55 - x}{2(55) - x - 37} \right) \times 10.$
$4 = \left( \frac{55 - x}{110 - x - 37} \right) \times 10 \implies 4 = \frac{550 - 10x}{73 - x}.$
$292 - 4x = 550 - 10x \implies 6x = 258 \implies x = 43.$
Substituting $x = 43$ in Equation $1$: $43 + y = 61 \implies y = 18.$
Thus,$x = 43$ and $y = 18.$

(A) Given total frequency $N = 60$.
Sum of frequencies: $5 + x + 20 + y + 2 = 60 \implies x + y + 27 = 60 \implies x + y = 33$ (Equation $1$).
Since the median is $50$,the median class is $40-60$. Here,$l = 40$,$h = 20$,$f = 20$,$cf = 5 + x$,and $N/2 = 30$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$50 = 40 + \left( \frac{30 - (5 + x)}{20} \right) \times 20$.
$50 = 40 + (25 - x)$.
$50 = 65 - x \implies x = 15$.
Substituting $x = 15$ in Equation $1$: $15 + y = 33 \implies y = 18$.
Thus,$x = 15$ and $y = 18$.

(A) Given total frequency $N = 100$,so $2 + 6 + 8 + x + 20 + 18 + y + 14 = 100$.
$68 + x + y = 100 \implies x + y = 32$ (Equation $1$).
Since the median is $49$,the median class is $40-50$. Here $l = 40$,$f = 20$,$cf = 2 + 6 + 8 + x = 16 + x$,and $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$.
$49 = 40 + \left( \frac{50 - (16 + x)}{20} \right) \times 10$.
$9 = \frac{34 - x}{2} \implies 18 = 34 - x \implies x = 16$.
Substituting $x = 16$ in Equation $1$: $16 + y = 32 \implies y = 16$.
Thus,$x = 16$ and $y = 16$.

(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) In the formula for calculating the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the term $l$ represents the lower limit of the modal class. The modal class is the class interval with the highest frequency.

(B) In the formula for the mode,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the terms are defined as follows:
$l$ is the lower limit of the modal class.
$f_{1}$ is the frequency of the modal class.
$f_{0}$ is the frequency of the class preceding the modal class.
$f_{2}$ is the frequency of the class succeeding the modal class.
$c$ is the class size or class length.
Therefore,$f_{0}$ represents the frequency of the class preceding the modal class.

(C) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the term $f_{1}$ represents the frequency of the modal class. Here,$l$ is the lower limit of the modal class,$f_{0}$ is the frequency of the class preceding the modal class,$f_{2}$ is the frequency of the class succeeding the modal class,and $c$ is the class size (or class length).

(D) In the formula for calculating the mode of grouped data,the variables are defined as follows:
$Z$ is the mode.
$l$ is the lower limit of the modal class.
$f_{1}$ is the frequency of the modal class.
$f_{0}$ is the frequency of the class preceding the modal class.
$f_{2}$ is the frequency of the class succeeding the modal class.
$c$ is the size of the class interval.
Therefore,$f_{2}$ represents the frequency of the class succeeding the modal class.

(A) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,the term $c$ represents the class size or class length of the modal class. The other terms are defined as follows: $l$ is the lower limit of the modal class,$f_{1}$ is the frequency of the modal class,$f_{0}$ is the frequency of the class preceding the modal class,and $f_{2}$ is the frequency of the class succeeding the modal class.

(B) In the formula for the median of grouped data,$M = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$,the term $l$ represents the lower limit of the median class.

(C) In the formula for the median,$M = l + \frac{(\frac{n}{2} - cf)}{f} \times h$,the term $n$ represents the total frequency of the given data distribution,which is the sum of all individual frequencies,i.e.,$n = \sum f_i$.