Mix Examples - Statistics Questions in English

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

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

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

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

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

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

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

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

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

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

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

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

(C) $1$. 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 $96$.

(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 $10 + 15 = 25$.

(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) = $145 - 125 = 20$.

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

(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
$0-20$	$17$
$20-40$	$5$
$40-60$	$7$
$60-80$	$8$
$80-100$	$13$

(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 $f_2 = 7$.

(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: $x = 38, y = 48$.

(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 $y = 1$.

(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 $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 50 + \left( \frac{12 - 7}{2(12) - 7 - 10} \right) \times 10 = 50 + \left( \frac{5}{24 - 17} \right) \times 10 = 50 + \frac{50}{7} \approx 50 + 7.14 = 57.14$.

(D) The empirical relationship between the mean $(\bar{x})$,median $(M)$,and mode $(Z)$ for a moderately skewed distribution is given by the formula: $Z = 3M - 2\bar{x}$.
Rearranging this formula:
$Z - M = (3M - 2\bar{x}) - M$
$Z - M = 2M - 2\bar{x}$
$Z - M = 2(M - \bar{x})$
Comparing this with the given expression $Z - M = \ldots \ldots \ldots \quad (M - \bar{x})$,the missing value is $2$.

(A) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:
$Z = 3M - 2\bar{x}$
Given values are $Z = 16$ and $M = 18$.
Substituting these values into the formula:
$16 = 3(18) - 2\bar{x}$
$16 = 54 - 2\bar{x}$
$2\bar{x} = 54 - 16$
$2\bar{x} = 38$
$\bar{x} = 19$
Therefore,the value of the mean is $19$.

(B) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given: $Z = 95$ and $\bar{x} = 98$.
Substituting the values into the formula:
$95 = 3M - 2(98)$
$95 = 3M - 196$
$3M = 95 + 196$
$3M = 291$
$M = \frac{291}{3} = 97$.
Therefore,the value of $M$ is $97$.

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:
$Z = 3M - 2\bar{x}$
Given:
$M = 62.5$
$\bar{x} = 64$
Substituting these values into the formula:
$Z = 3(62.5) - 2(64)$
$Z = 187.5 - 128$
$Z = 59.5$
Therefore,the value of $Z$ is $59.5$.

(D) Given equations are:
$M + \bar{x} = 32$ --- $(1)$
$M - \bar{x} = 2$ --- $(2)$
Adding equation $(1)$ and $(2)$:
$(M + \bar{x}) + (M - \bar{x}) = 32 + 2$
$2M = 34$
$M = 17$
Substituting $M = 17$ in equation $(1)$:
$17 + \bar{x} = 32$
$\bar{x} = 32 - 17 = 15$
Using the empirical relationship between Mean,Median,and Mode:
$Z = 3M - 2\bar{x}$
Substituting the values of $M$ and $\bar{x}$:
$Z = 3(17) - 2(15)$
$Z = 51 - 30$
$Z = 21$
Therefore,the value of $Z$ is $21$.

(A) Given equations are:
$Z + M = 108$ --- $(1)$
$Z - M = -6$ --- $(2)$
Adding $(1)$ and $(2)$:
$(Z + M) + (Z - M) = 108 - 6$
$2Z = 102$
$Z = 51$
Substituting $Z = 51$ in $(1)$:
$51 + M = 108$
$M = 108 - 51 = 57$
Using the empirical relationship between Mean,Median,and Mode:
$Mode = 3 \times Median - 2 \times Mean$
$Z = 3M - 2\bar{x}$
$51 = 3(57) - 2\bar{x}$
$51 = 171 - 2\bar{x}$
$2\bar{x} = 171 - 51$
$2\bar{x} = 120$
$\bar{x} = 60$

(B) Given equations are:
$(1)$ $\bar{x} + Z = 37.5$
$(2)$ $\bar{x} - Z = 1.5$
Adding $(1)$ and $(2)$: $2\bar{x} = 39 \implies \bar{x} = 19.5$
Subtracting $(2)$ from $(1)$: $2Z = 36 \implies Z = 18$
Using the empirical relationship between Mean,Median,and Mode: $Z = 3M - 2\bar{x}$
$18 = 3M - 2(19.5)$
$18 = 3M - 39$
$3M = 18 + 39 = 57$
$M = 57 / 3 = 19$
Therefore,the median $M$ is $19$.

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given,$Z - M = 2$,which implies $M = Z - 2$.
Also,$\bar{x} = 33.5$.
Substituting these values into the formula:
$Z = 3(Z - 2) - 2(33.5)$
$Z = 3Z - 6 - 67$
$Z = 3Z - 73$
$73 = 3Z - Z$
$73 = 2Z$
$Z = 36.5$.

(D) The empirical relationship between Mean $(\bar{x})$, Median $(M)$, and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given values are $M = 15.4$ and $\bar{x} = 14.5$.
Substituting these values into the formula:
$Z = 3(15.4) - 2(14.5)$
$Z = 46.2 - 29.0$
$Z = 17.2$.
Therefore, the value of $Z$ is $17.2$.

(A) Given equations are:
$(1)$ $M + \bar{x} = 165$
$(2)$ $M - \bar{x} = 1$
Adding equations $(1)$ and $(2)$:
$2M = 166 \implies M = 83$
Subtracting equation $(2)$ from $(1)$:
$2\bar{x} = 164 \implies \bar{x} = 82$
Using the empirical relationship between Mean,Median,and Mode:
$Z = 3M - 2\bar{x}$
$Z = 3(83) - 2(82)$
$Z = 249 - 164$
$Z = 85$

(B) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given that $Z - M = 12$,we can write $Z = M + 12$.
Substituting this into the empirical formula:
$M + 12 = 3M - 2\bar{x}$
$12 = 3M - M - 2\bar{x}$
$12 = 2M - 2\bar{x}$
$12 = 2(M - \bar{x})$
Dividing both sides by $2$:
$M - \bar{x} = 6$.

(D) In an exclusive class interval (or continuous series),the lower limit is included in the class,but the upper limit is excluded.
For the class interval $30-45$:
- The lower limit is $30$,which is included in this class.
- The upper limit is $45$,which is excluded from this class and is instead included in the next class interval (e.g.,$45-60$).
Therefore,the value $45$ is not included in the class $30-45$.

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:
$Z = 3M - 2\bar{x}$
Given values are $Z = 12$ and $M = 16$.
Substituting these values into the formula:
$12 = 3(16) - 2\bar{x}$
$12 = 48 - 2\bar{x}$
Rearranging the equation to solve for $\bar{x}$:
$2\bar{x} = 48 - 12$
$2\bar{x} = 36$
$\bar{x} = \frac{36}{2}$
$\bar{x} = 18$

(D) The empirical relationship between mean,median,and mode is given by the formula:
$Z = 3M - 2\bar{x}$
where $Z$ is the mode,$M$ is the median,and $\bar{x}$ is the mean.
Given: $Z = 25$ and $\bar{x} = 19$.
Substituting these values into the formula:
$25 = 3M - 2(19)$
$25 = 3M - 38$
$3M = 25 + 38$
$3M = 63$
$M = \frac{63}{3} = 21$
Therefore,the median is $21$.

(B) The empirical relationship between Mean,Median,and Mode is given by the formula:
$Z = 3M - 2\bar{x}$
Given that $M = 72$ and $\bar{x} = 70$,we substitute these values into the formula:
$Z = 3(72) - 2(70)$
$Z = 216 - 140$
$Z = 76$
Therefore,the value of $Z$ is $76$.

(A) We know the empirical relationship between mean,median,and mode is given by: $Z = 3M - 2\bar{x}$.
We can rewrite this equation as:
$Z - M = 3M - 2\bar{x} - M$
$Z - M = 2M - 2\bar{x}$
$Z - M = 2(M - \bar{x})$
Given that $Z - M = 4$,we substitute this value into the equation:
$4 = 2(M - \bar{x})$
Dividing both sides by $2$:
$M - \bar{x} = \frac{4}{2}$
$M - \bar{x} = 2$

(C) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula:
$Z = 3M - 2\bar{x}$
Given values are $Z = 36.8$ and $M = 33.6$.
Substituting these values into the formula:
$36.8 = 3(33.6) - 2\bar{x}$
Calculate $3 \times 33.6$:
$36.8 = 100.8 - 2\bar{x}$
Rearrange the equation to solve for $\bar{x}$:
$2\bar{x} = 100.8 - 36.8$
$2\bar{x} = 64$
Divide by $2$:
$\bar{x} = \frac{64}{2} = 32$
Thus,the value of $\bar{x}$ is $32$.

(C) We know the empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by: $Z = 3M - 2\bar{x}$.
Given equations:
$1) Z + M = 34 \implies Z = 34 - M$
$2) M + \bar{x} = 40 \implies \bar{x} = 40 - M$
Substitute these into the empirical formula:
$34 - M = 3M - 2(40 - M)$
$34 - M = 3M - 80 + 2M$
$34 - M = 5M - 80$
$34 + 80 = 5M + M$
$114 = 6M$
$M = \frac{114}{6} = 19$
Therefore,the value of $M$ is $19$.

(B) The empirical relationship between Mean $( \bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given that $\bar{x} = Z - 3$,we can express the mode as $Z = \bar{x} + 3$.
Substituting these values into the empirical formula:
$\bar{x} + 3 = 3(22) - 2\bar{x}$
$\bar{x} + 3 = 66 - 2\bar{x}$
Adding $2\bar{x}$ to both sides:
$3\bar{x} + 3 = 66$
Subtracting $3$ from both sides:
$3\bar{x} = 63$
Dividing by $3$:
$\bar{x} = 21$.
Therefore,the mean is $21$.

(A) The empirical relationship between Mean $( \bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
Given that $M - \bar{x} = 2$,we can express the mean as $\bar{x} = M - 2$.
Substituting this into the empirical formula:
$Z = 3M - 2(M - 2)$
$20.5 = 3M - 2M + 4$
$20.5 = M + 4$
$M = 20.5 - 4$
$M = 16.5$
Therefore,the median $M$ is $16.5$.

(C) Given the relation $3 \bar{x} = 2 M = 60.$
From this,we can find the values of mean $(\bar{x})$ and median $(M)$:
$\bar{x} = \frac{60}{3} = 20$
$M = \frac{60}{2} = 30$
The empirical relationship between mean,median,and mode $(Z)$ is given by the formula:
$Z = 3 M - 2 \bar{x}$
Substituting the values of $M$ and $\bar{x}$ into the formula:
$Z = 3(30) - 2(20)$
$Z = 90 - 40$
$Z = 50$

(B) The empirical relationship between mean,median,and mode is given by the formula: $Mode = 3 \times \text{Median} - 2 \times \text{Mean}$.
Given:
Mean $(\bar{x})$ = $72.5$
Median $(M)$ = $73.9$
Substituting these values into the formula:
$Mode = 3(73.9) - 2(72.5)$
$Mode = 221.7 - 145.0$
$Mode = 76.7$
Therefore,the mode of the data is $76.7$.

(D) The empirical relationship between Mean $(\bar{x})$,Median $(M)$,and Mode $(Z)$ is given by the formula: $Z = 3M - 2\bar{x}$.
To find the value of the missing term in $Z - M = \ldots \times (M - \bar{x})$,we rearrange the formula:
$Z = 3M - 2\bar{x}$
Subtract $M$ from both sides:
$Z - M = 3M - 2\bar{x} - M$
$Z - M = 2M - 2\bar{x}$
Factor out $2$:
$Z - M = 2(M - \bar{x})$
Comparing this with the given expression,the missing value is $2$.