Given: A frequency distribution table representing the monthly electricity consumption (in units) of 68 consumers.

Monthly Consumption (units)	Number of Consumers ($f_i$)
65 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

To Find: The Mean, Median, and Mode of the given data and compare them.

Step 1: Calculation of Mean ($\bar{x}$) using the Step-Deviation Method

Let the assumed mean $a = 135$. The class size $h = 20$.

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$d_i = x_i - 135$	$u_i = \frac{d_i}{20}$	$f_i u_i$
65-85	4	75	-60	-3	-12
85-105	5	95	-40	-2	-10
105-125	13	115	-20	-1	-13
125-145	20	135	0	0	0
145-165	14	155	20	1	14
165-185	8	175	40	2	16
185-205	4	195	60	3	12
Total	$\sum f_i = 68$	-	-	-	$\sum f_i u_i = 7$

Formula for Mean: $\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$

$\bar{x} = 135 + \left( \frac{7}{68} \right) \times 20 = 135 + \frac{140}{68} \approx 135 + 2.06 = 137.06$

Step 2: Calculation of Median

Cumulative Frequency ($cf$) table:

Class Interval	Frequency ($f_i$)	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, $n = 68$, so $\frac{n}{2} = 34$. The cumulative frequency just greater than 34 is 42, which corresponds to the class 125-145.

Median Class = 125-145. Lower limit ($l$) = 125, $f = 20$, $cf$ of preceding class = 22, $h = 20$.

Median = $l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h = 125 + \left( \frac{34 - 22}{20} \right) \times 20 = 125 + 12 = 137$.

Step 3: Calculation of Mode

The maximum frequency is 20, which corresponds to the class 125-145. This is the Modal Class.

$l = 125, f_1 = 20, f_0 = 13, f_2 = 14, h = 20$.

Mode = $l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 125 + \left( \frac{20 - 13}{2(20) - 13 - 14} \right) \times 20$

Mode = $125 + \left( \frac{7}{40 - 27} \right) \times 20 = 125 + \left( \frac{7}{13} \right) \times 20 = 125 + 10.77 = 135.77$.

Step 4: Comparison

Mean $\approx 137.06$, Median $= 137$, Mode $\approx 135.77$.

The values are very close to each other, indicating a nearly symmetric distribution.

Final Answer: Mean = 137.06 units, Median = 137 units, Mode = 135.77 units.

Given: The frequency distribution of ages of 100 policy holders, provided as a cumulative frequency table (less than type):

Age (years)	Number of policy holders (Cumulative Frequency)
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

To Find: The median age of the policy holders.

Step 1: Convert the cumulative frequency distribution into a standard frequency distribution table.

To calculate the median, we need the class intervals and their corresponding frequencies ($f_i$).

Class Interval	Frequency ($f_i$)	Cumulative Frequency ($cf$)
15-20	2	2
20-25	6 - 2 = 4	6
25-30	24 - 6 = 18	24
30-35	45 - 24 = 21	45
35-40	78 - 45 = 33	78
40-45	89 - 78 = 11	89
45-50	92 - 89 = 3	92
50-55	98 - 92 = 6	98
55-60	100 - 98 = 2	100
Total	$N = 100$	-

Step 2: Identify the median class.

The total number of observations is $N = 100$.

We calculate $\frac{N}{2} = \frac{100}{2} = 50$.

Looking at the cumulative frequency column, the first cumulative frequency greater than 50 is 78, which corresponds to the class interval 35-40.

Therefore, the median class is 35-40.

Step 3: Apply the Median Formula.

The formula for the median of a grouped frequency distribution is:

$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$

Where:

• $l$ (lower limit of median class) = $35$
• $N$ (total frequency) = $100$
• $cf$ (cumulative frequency of the class preceding the median class) = $45$
• $f$ (frequency of the median class) = $33$
• $h$ (class size) = $40 - 35 = 5$

Step 4: Perform the calculation.

Substitute the values into the formula:

$\text{Median} = 35 + \left( \frac{50 - 45}{33} \right) \times 5$

$\text{Median} = 35 + \left( \frac{5}{33} \right) \times 5$

$\text{Median} = 35 + \frac{25}{33}$

$\text{Median} = 35 + 0.7575...$

$\text{Median} \approx 35.76$

Final Answer: The median age of the policy holders is 35.76 years.

Given: A frequency distribution table with a total frequency $N = 60$ and a median value of $28.5$.

Class Interval	Frequency ($f$)
0-10	5
10-20	$x$
20-30	20
30-40	15
40-50	$y$
50-60	5
Total	60

To find: The values of $x$ and $y$.

Step 1: Constructing the Cumulative Frequency Table

Class Interval	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	$45 + x + y$

Step 2: Establishing the first linear equation

Since the total frequency is given as $60$, we have:

$45 + x + y = 60$
$x + y = 60 - 45$
$x + y = 15$ --- (Equation 1)

Step 3: Identifying the Median Class

The median is given as $28.5$. Since $28.5$ lies between $20$ and $30$, the median class is $20-30$.

From the median class $20-30$:

• Lower limit ($l$) = $20$
• Frequency of median class ($f$) = $20$
• Cumulative frequency of the class preceding the median class ($cf$) = $5 + x$
• Class size ($h$) = $10$
• $N/2 = 60/2 = 30$

Step 4: Applying the Median Formula

The formula for the median is: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$

Substituting the known values:

$28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10$
$28.5 - 20 = \left( \frac{30 - 5 - x}{20} \right) \times 10$
$8.5 = \frac{25 - x}{2}$
$8.5 \times 2 = 25 - x$
$17 = 25 - x$
$x = 25 - 17$
$x = 8$

Step 5: Solving for $y$

Substitute $x = 8$ into Equation 1:

$8 + y = 15$
$y = 15 - 8$
$y = 7$

Final Answer: The values are $x = 8$ and $y = 7$.

Given: The frequency distribution of the lengths of 40 leaves measured to the nearest millimetre.

Length (mm)	Number of leaves ($f$)
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

To Find: The median length of the leaves.

Step 1: Converting to Continuous Classes
The given classes are discontinuous (e.g., 126 and 127). To make them continuous, we subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class interval.

Class Interval	Frequency ($f$)	Cumulative Frequency ($cf$)
117.5 - 126.5	3	3
126.5 - 135.5	5	8
135.5 - 144.5	9	17
144.5 - 153.5	12	29
153.5 - 162.5	5	34
162.5 - 171.5	4	38
171.5 - 180.5	2	40
Total	$n = 40$	-

Step 2: Identifying the Median Class
The total number of observations is $n = 40$.
We calculate $\frac{n}{2} = \frac{40}{2} = 20$.
Looking at the cumulative frequency column, the first cumulative frequency greater than 20 is 29, which corresponds to the class interval $144.5 - 153.5$.
Thus, the median class is $144.5 - 153.5$.

Step 3: Applying the Median Formula
The formula for the median is:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
Where:
$l$ (lower limit of median class) = $144.5$
$n$ (total number of observations) = $40$
$cf$ (cumulative frequency of the class preceding the median class) = $17$
$f$ (frequency of the median class) = $12$
$h$ (class size) = $153.5 - 144.5 = 9$

Step 4: Calculation
$\text{Median} = 144.5 + \left( \frac{20 - 17}{12} \right) \times 9$
$\text{Median} = 144.5 + \left( \frac{3}{12} \right) \times 9$
$\text{Median} = 144.5 + \left( \frac{1}{4} \right) \times 9$
$\text{Median} = 144.5 + 2.25$
$\text{Median} = 146.75$

Final Answer: The median length of the leaves is 146.75 mm.

Given: The frequency distribution of the lifetime of 400 neon lamps as follows:

Lifetime (hours)	Number of Lamps ($f_i$)
1500 - 2000	14
2000 - 2500	56
2500 - 3000	60
3000 - 3500	86
3500 - 4000	74
4000 - 4500	62
4500 - 5000	48

To Find: The median lifetime of the neon lamps.

Step 1: Construct the Cumulative Frequency Table

To find the median, we first calculate the cumulative frequency ($cf$) for each class interval.

Lifetime (hours)	Frequency ($f_i$)	Cumulative Frequency ($cf$)
1500 - 2000	14	14
2000 - 2500	56	70
2500 - 3000	60	130
3000 - 3500	86	216
3500 - 4000	74	290
4000 - 4500	62	352
4500 - 5000	48	400
Total	$N = 400$	-

Step 2: Identify the Median Class

The total number of observations is $N = 400$.
We calculate $\frac{N}{2} = \frac{400}{2} = 200$.
Looking at the cumulative frequency column, the first cumulative frequency greater than 200 is 216, which corresponds to the class interval $3000 - 3500$.
Therefore, the median class is $3000 - 3500$.

Step 3: Apply the Median Formula

The formula for the median of a grouped frequency distribution is:
$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
Where:
$l$ (lower limit of median class) = $3000$
$N$ (total frequency) = $400$
$cf$ (cumulative frequency of the class preceding the median class) = $130$
$f$ (frequency of the median class) = $86$
$h$ (class size) = $500$

Step 4: Perform the Calculation

Substitute the values into the formula:
$\text{Median} = 3000 + \left( \frac{200 - 130}{86} \right) \times 500$
$\text{Median} = 3000 + \left( \frac{70}{86} \right) \times 500$
$\text{Median} = 3000 + \left( \frac{35000}{86} \right)$ [Simplifying the fraction: $\frac{70}{86} = \frac{35}{43}$]
$\text{Median} = 3000 + 406.9767...$
$\text{Median} \approx 3406.98$

Final Answer: The median lifetime of a lamp is 3406.98 hours.

Given: A frequency distribution of the weights of 30 students.

Weight (kg)	Number of students (f)
40 - 45	2
45 - 50	3
50 - 55	8
55 - 60	6
60 - 65	6
65 - 70	3
70 - 75	2

To Find: The median weight of the students.

Step 1: Constructing the Cumulative Frequency Table

To find the median, we must first calculate the cumulative frequency ($cf$) for each class interval.

Weight (kg)	Frequency ($f$)	Cumulative Frequency ($cf$)
40 - 45	2	2
45 - 50	3	2 + 3 = 5
50 - 55	8	5 + 8 = 13
55 - 60	6	13 + 6 = 19
60 - 65	6	19 + 6 = 25
65 - 70	3	25 + 3 = 28
70 - 75	2	28 + 2 = 30
Total	$n = 30$	-

Step 2: Identifying the Median Class

The total number of observations is $n = 30$.
We calculate $\frac{n}{2} = \frac{30}{2} = 15$.
[The median class is the class interval whose cumulative frequency is greater than or nearest to $\frac{n}{2}$.]
Looking at the table, the cumulative frequency just greater than 15 is 19, which corresponds to the class interval 55 - 60.

Step 3: Applying the Median Formula

The formula for the median is:
$Median = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
Where:
$l$ (lower limit of median class) = $55$
$n$ (total number of observations) = $30$
$cf$ (cumulative frequency of the class preceding the median class) = $13$
$f$ (frequency of the median class) = $6$
$h$ (class size) = $60 - 55 = 5$

Step 4: Calculating the Result

$Median = 55 + \left( \frac{15 - 13}{6} \right) \times 5$
$Median = 55 + \left( \frac{2}{6} \right) \times 5$
$Median = 55 + \left( \frac{1}{3} \right) \times 5$
$Median = 55 + \frac{5}{3}$
$Median = 55 + 1.666...$
$Median \approx 56.67$

Final Answer: The median weight of the students is 56.67 kg.

Given: A frequency distribution of the number of letters in 100 surnames.

Number of letters	Number of surnames ($f_i$)
1 - 4	6
4 - 7	30
7 - 10	40
10 - 13	16
13 - 16	4
16 - 19	4

To Find: The Median, Mean, and Mode of the given data.

Step 1: Calculation of Mean ($\bar{x}$)

We use the Assumed Mean Method. Let the assumed mean $a = 11.5$. The class size $h = 3$.

Class Interval	Frequency ($f_i$)	Class Mark ($x_i$)	$d_i = x_i - a$	$u_i = \frac{x_i - a}{h}$	$f_i u_i$
1-4	6	2.5	-9	-3	-18
4-7	30	5.5	-6	-2	-60
7-10	40	8.5	-3	-1	-40
10-13	16	11.5	0	0	0
13-16	4	14.5	3	1	4
16-19	4	17.5	6	2	8
Total	$\sum f_i = 100$	-	-	-	$\sum f_i u_i = -106$

Formula for Mean: $\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$

$\bar{x} = 11.5 + \left( \frac{-106}{100} \right) \times 3 = 11.5 - 3.18 = 8.32$

Step 2: Calculation of Median

Total frequency $N = 100$. So, $N/2 = 50$.

Class Interval	Frequency ($f_i$)	Cumulative Frequency ($cf$)
1-4	6	6
4-7	30	36
7-10	40	76

The cumulative frequency just greater than 50 is 76, so the median class is $7-10$.

Lower limit ($l$) = 7, $cf$ of preceding class = 36, $f$ of median class = 40, $h = 3$.

Median = $l + \left( \frac{N/2 - cf}{f} \right) \times h = 7 + \left( \frac{50 - 36}{40} \right) \times 3 = 7 + \left( \frac{14}{40} \right) \times 3 = 7 + 1.05 = 8.05$

Step 3: Calculation of Mode

The modal class is the class with the highest frequency, which is $7-10$ ($f_1 = 40$).

$l = 7, f_1 = 40, f_0 = 30, f_2 = 16, h = 3$.

Mode = $l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 7 + \left( \frac{40 - 30}{2(40) - 30 - 16} \right) \times 3$

Mode = $7 + \left( \frac{10}{80 - 46} \right) \times 3 = 7 + \left( \frac{10}{34} \right) \times 3 = 7 + 0.88 = 7.88$

Final Answer: The Mean number of letters is 8.32, the Median is 8.05, and the Mode is 7.88.