Statistics

Given: Frequency distribution of number of plants in 20 houses.

Method Used: Direct Method (since the values of $x_i$ are small and easy to compute).

Formula: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

Step 1: Find the midpoint $x_i$ of each class.

| Class | $f_i$ | $x_i$ (midpoint) | $f_i x_i$ |
|---|---|---|---|
| 0 – 2 | 1 | 1 | 1 |
| 2 – 4 | 2 | 3 | 6 |
| 4 – 6 | 1 | 5 | 5 |
| 6 – 8 | 5 | 7 | 35 |
| 8 – 10 | 6 | 9 | 54 |
| 10 – 12 | 2 | 11 | 22 |
| 12 – 14 | 3 | 13 | 39 |
| Total | 20 | | 162 |

Step 2: Calculate the mean.
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1$$

Answer: The mean number of plants per house is 8.1.

Reason for choosing Direct Method: The class midpoints $x_i$ are small integers, making direct multiplication simple without needing an assumed mean.

Given: Frequency distribution of daily wages of 50 workers.

Method Used: Assumed Mean Method (since the values of $x_i$ are large).

Let assumed mean $a = 550$, class width $h = 20$.

Formula: $$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad \text{where } d_i = x_i - a$$

Step 1: Prepare the table.

| Class | $f_i$ | $x_i$ | $d_i = x_i - 550$ | $f_i d_i$ |
|---|---|---|---|---|
| 500 – 520 | 12 | 510 | –40 | –480 |
| 520 – 540 | 14 | 530 | –20 | –280 |
| 540 – 560 | 8 | 550 | 0 | 0 |
| 560 – 580 | 6 | 570 | 20 | 120 |
| 580 – 600 | 10 | 590 | 40 | 400 |
| Total | 50 | | | –240 |

Step 2: Calculate the mean.
$$\bar{x} = 550 + \frac{-240}{50} = 550 - 4.8 = 545.20$$

Answer: The mean daily wages of the workers is ₹ 545.20.

Given: Mean $\bar{x} = 18$, one frequency is missing.

Method: Direct Method.

Formula: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

Step 1: Prepare the table.

| Class | $f_i$ | $x_i$ (midpoint) | $f_i x_i$ |
|---|---|---|---|
| 11 – 13 | 7 | 12 | 84 |
| 13 – 15 | 6 | 14 | 84 |
| 15 – 17 | 9 | 16 | 144 |
| 17 – 19 | 13 | 18 | 234 |
| 19 – 21 | $f$ | 20 | $20f$ |
| 21 – 23 | 5 | 22 | 110 |
| 23 – 25 | 4 | 24 | 96 |
| Total | $44 + f$ | | $752 + 20f$ |

Step 2: Apply the mean formula.
$$18 = \frac{752 + 20f}{44 + f}$$

$$18(44 + f) = 752 + 20f$$

$$792 + 18f = 752 + 20f$$

$$792 - 752 = 20f - 18f$$

$$40 = 2f$$

$$f = 20$$

Answer: The missing frequency $f$ = 20.

Given: Frequency distribution of heartbeats per minute for 30 women.

Method Used: Assumed Mean Method.

Let assumed mean $a = 75.5$, class width $h = 3$.

Formula: $$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad d_i = x_i - a$$

Step 1: Prepare the table.

| Class | $f_i$ | $x_i$ | $d_i = x_i - 75.5$ | $f_i d_i$ |
|---|---|---|---|---|
| 65 – 68 | 2 | 66.5 | –9 | –18 |
| 68 – 71 | 4 | 69.5 | –6 | –24 |
| 71 – 74 | 3 | 72.5 | –3 | –9 |
| 74 – 77 | 8 | 75.5 | 0 | 0 |
| 77 – 80 | 7 | 78.5 | 3 | 21 |
| 80 – 83 | 4 | 81.5 | 6 | 24 |
| 83 – 86 | 2 | 84.5 | 9 | 18 |
| Total | 30 | | | 12 |

Step 2: Calculate the mean.
$$\bar{x} = 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$$

Answer: The mean heartbeats per minute for these women is 75.9 beats per minute.

Given: The classes are not continuous (gaps exist: 50–52, 53–55, …). We first note that these are inclusive classes. The midpoints are taken directly.

Method Used: Step Deviation Method (since frequencies are large and values are close together).

Let assumed mean $a = 57$, class width $h = 3$.

Formula: $$\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h, \quad u_i = \frac{x_i - a}{h}$$

Step 1: Find midpoints and prepare the table.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i-57}{3}$ | $f_i u_i$ |
|---|---|---|---|---|
| 50 – 52 | 15 | 51 | –2 | –30 |
| 53 – 55 | 110 | 54 | –1 | –110 |
| 56 – 58 | 135 | 57 | 0 | 0 |
| 59 – 61 | 115 | 60 | 1 | 115 |
| 62 – 64 | 25 | 63 | 2 | 50 |
| Total | 400 | | | 25 |

Step 2: Calculate the mean.
$$\bar{x} = 57 + \frac{25}{400} \times 3 = 57 + \frac{75}{400} = 57 + 0.1875 = 57.19 \text{ (approx.)}$$

Answer: The mean number of mangoes kept in a packing box is 57.19 (approximately).

Reason: Step deviation method was chosen because the frequencies are large and the class widths are equal, making calculations simpler.

Given: Frequency distribution of daily food expenditure of 25 households.

Method Used: Step Deviation Method.

Let assumed mean $a = 225$, class width $h = 50$.

Formula: $$\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h, \quad u_i = \frac{x_i - a}{h}$$

Step 1: Prepare the table.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i - 225}{50}$ | $f_i u_i$ |
|---|---|---|---|---|
| 100 – 150 | 4 | 125 | –2 | –8 |
| 150 – 200 | 5 | 175 | –1 | –5 |
| 200 – 250 | 12 | 225 | 0 | 0 |
| 250 – 300 | 2 | 275 | 1 | 2 |
| 300 – 350 | 2 | 325 | 2 | 4 |
| Total | 25 | | | –7 |

Step 2: Calculate the mean.
$$\bar{x} = 225 + \frac{-7}{25} \times 50 = 225 - 14 = 211$$

Answer: The mean daily expenditure on food is ₹ 211.

Given: Frequency distribution of SO₂ concentration for 30 localities.

Method Used: Direct Method.

Formula: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

Step 1: Find midpoints and prepare the table.

| Class | $f_i$ | $x_i$ (midpoint) | $f_i x_i$ |
|---|---|---|---|
| 0.00 – 0.04 | 4 | 0.02 | 0.08 |
| 0.04 – 0.08 | 9 | 0.06 | 0.54 |
| 0.08 – 0.12 | 9 | 0.10 | 0.90 |
| 0.12 – 0.16 | 2 | 0.14 | 0.28 |
| 0.16 – 0.20 | 4 | 0.18 | 0.72 |
| 0.20 – 0.24 | 2 | 0.22 | 0.44 |
| Total | 30 | | 2.96 |

Step 2: Calculate the mean.
$$\bar{x} = \frac{2.96}{30} = 0.0987 \text{ ppm (approx.)}$$

Answer: The mean concentration of SO₂ in the air is 0.099 ppm (approximately).

Given: Frequency distribution of absentee days for 40 students.

Method Used: Direct Method.

Formula: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

Step 1: Find midpoints and prepare the table.

| Class | $f_i$ | $x_i$ (midpoint) | $f_i x_i$ |
|---|---|---|---|
| 0 – 6 | 11 | 3 | 33 |
| 6 – 10 | 10 | 8 | 80 |
| 10 – 14 | 7 | 12 | 84 |
| 14 – 20 | 4 | 17 | 68 |
| 20 – 28 | 4 | 24 | 96 |
| 28 – 38 | 3 | 33 | 99 |
| 38 – 40 | 1 | 39 | 39 |
| Total | 40 | | 499 |

Step 2: Calculate the mean.
$$\bar{x} = \frac{499}{40} = 12.475$$

Answer: The mean number of days a student was absent is 12.48 days (approximately).

Given: Frequency distribution of literacy rates of 35 cities.

Method Used: Step Deviation Method.

Let assumed mean $a = 70$, class width $h = 10$.

Formula: $$\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h, \quad u_i = \frac{x_i - a}{h}$$

Step 1: Prepare the table.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i - 70}{10}$ | $f_i u_i$ |
|---|---|---|---|---|
| 45 – 55 | 3 | 50 | –2 | –6 |
| 55 – 65 | 10 | 60 | –1 | –10 |
| 65 – 75 | 11 | 70 | 0 | 0 |
| 75 – 85 | 8 | 80 | 1 | 8 |
| 85 – 95 | 3 | 90 | 2 | 6 |
| Total | 35 | | | –2 |

Step 2: Calculate the mean.
$$\bar{x} = 70 + \frac{-2}{35} \times 10 = 70 - \frac{20}{35} = 70 - \frac{4}{7} = 70 - 0.571 = 69.43 \text{ (approx.)}$$

Answer: The mean literacy rate is 69.43% (approximately).

Given: Frequency distribution of ages of patients.

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FINDING MODE:

Step 1: Identify the modal class (class with highest frequency).

Highest frequency = 23, which belongs to class 35 – 45.

So, Modal class = 35 – 45.

$l = 35$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$

Formula: $$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

$$\text{Mode} = 35 + \frac{23 - 21}{2(23) - 21 - 14} \times 10 = 35 + \frac{2}{46 - 35} \times 10 = 35 + \frac{2}{11} \times 10$$

$$\text{Mode} = 35 + \frac{20}{11} = 35 + 1.81 = 36.8 \text{ years (approx.)}$$

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FINDING MEAN:

Let assumed mean $a = 30$, $h = 10$.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i-30}{10}$ | $f_i u_i$ |
|---|---|---|---|---|
| 5 – 15 | 6 | 10 | –2 | –12 |
| 15 – 25 | 11 | 20 | –1 | –11 |
| 25 – 35 | 21 | 30 | 0 | 0 |
| 35 – 45 | 23 | 40 | 1 | 23 |
| 45 – 55 | 14 | 50 | 2 | 28 |
| 55 – 65 | 5 | 60 | 3 | 15 |
| Total | 80 | | | 43 |

$$\bar{x} = 30 + \frac{43}{80} \times 10 = 30 + 5.375 = 35.375 \approx 35.38 \text{ years}$$

---

Interpretation:
- Mean age ≈ 35.38 years — this is the average age of all patients.
- Mode ≈ 36.8 years — this is the age group most commonly admitted.

Both values are close, indicating that the maximum number of patients admitted are in the age group around 35–45 years.

Given: Frequency distribution of lifetimes of 225 electrical components.

Step 1: Identify the modal class.

Highest frequency = 61, which belongs to class 60 – 80.

Modal class = 60 – 80.

$l = 60$, $f_1 = 61$, $f_0 = 52$, $f_2 = 38$, $h = 20$

Formula: $$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

$$\text{Mode} = 60 + \frac{61 - 52}{2(61) - 52 - 38} \times 20$$

$$= 60 + \frac{9}{122 - 90} \times 20 = 60 + \frac{9}{32} \times 20$$

$$= 60 + \frac{180}{32} = 60 + 5.625 = 65.625 \text{ hours}$$

Answer: The modal lifetime of the components is 65.625 hours.

Given: Frequency distribution of monthly expenditure of 200 families.

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FINDING MODE:

Highest frequency = 40, Modal class = 1500 – 2000.

$l = 1500$, $f_1 = 40$, $f_0 = 24$, $f_2 = 33$, $h = 500$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

$$= 1500 + \frac{40 - 24}{2(40) - 24 - 33} \times 500 = 1500 + \frac{16}{80 - 57} \times 500$$

$$= 1500 + \frac{16}{23} \times 500 = 1500 + \frac{8000}{23} = 1500 + 347.83 = 1847.83$$

Modal monthly expenditure ≈ ₹ 1847.83

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FINDING MEAN:

Let assumed mean $a = 3250$, $h = 500$.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i - 3250}{500}$ | $f_i u_i$ |
|---|---|---|---|---|
| 1000 – 1500 | 24 | 1250 | –4 | –96 |
| 1500 – 2000 | 40 | 1750 | –3 | –120 |
| 2000 – 2500 | 33 | 2250 | –2 | –66 |
| 2500 – 3000 | 28 | 2750 | –1 | –28 |
| 3000 – 3500 | 30 | 3250 | 0 | 0 |
| 3500 – 4000 | 22 | 3750 | 1 | 22 |
| 4000 – 4500 | 16 | 4250 | 2 | 32 |
| 4500 – 5000 | 7 | 4750 | 3 | 21 |
| Total | 200 | | | –235 |

$$\bar{x} = 3250 + \frac{-235}{200} \times 500 = 3250 - \frac{235 \times 500}{200} = 3250 - 587.5 = 2662.5$$

Answer: Modal monthly expenditure = ₹ 1847.83 and Mean monthly expenditure = ₹ 2662.50.

Given: Frequency distribution of teacher-student ratio.

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FINDING MODE:

Highest frequency = 10, Modal class = 30 – 35.

$l = 30$, $f_1 = 10$, $f_0 = 9$, $f_2 = 3$, $h = 5$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 30 + \frac{10 - 9}{2(10) - 9 - 3} \times 5$$

$$= 30 + \frac{1}{20 - 12} \times 5 = 30 + \frac{5}{8} = 30 + 0.625 = 30.6 \text{ (approx.)}$$

---

FINDING MEAN:

Let assumed mean $a = 32.5$, $h = 5$.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i - 32.5}{5}$ | $f_i u_i$ |
|---|---|---|---|---|
| 15 – 20 | 3 | 17.5 | –3 | –9 |
| 20 – 25 | 8 | 22.5 | –2 | –16 |
| 25 – 30 | 9 | 27.5 | –1 | –9 |
| 30 – 35 | 10 | 32.5 | 0 | 0 |
| 35 – 40 | 3 | 37.5 | 1 | 3 |
| 40 – 45 | 0 | 42.5 | 2 | 0 |
| 45 – 50 | 0 | 47.5 | 3 | 0 |
| 50 – 55 | 2 | 52.5 | 4 | 8 |
| Total | 35 | | | –23 |

$$\bar{x} = 32.5 + \frac{-23}{35} \times 5 = 32.5 - \frac{115}{35} = 32.5 - 3.286 = 29.2 \text{ (approx.)}$$

---

Interpretation:
- Mode ≈ 30.6 students per teacher — the most common teacher-student ratio among states/UTs.
- Mean ≈ 29.2 students per teacher — the average teacher-student ratio across all states/UTs.

The mode is slightly higher than the mean, indicating that more states have a ratio around 30–35 students per teacher.

Given: Frequency distribution of runs scored by top batsmen.

Step 1: Identify the modal class.

Highest frequency = 18, Modal class = 4000 – 5000.

$l = 4000$, $f_1 = 18$, $f_0 = 4$, $f_2 = 9$, $h = 1000$

Formula: $$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

$$= 4000 + \frac{18 - 4}{2(18) - 4 - 9} \times 1000 = 4000 + \frac{14}{36 - 13} \times 1000$$

$$= 4000 + \frac{14}{23} \times 1000 = 4000 + 608.7 = 4608.7 \text{ runs (approx.)}$$

Answer: The mode of the data is 4608.7 runs (approximately).

Given: Frequency distribution of number of cars passing through a spot.

Step 1: Identify the modal class.

Highest frequency = 20, Modal class = 40 – 50.

$l = 40$, $f_1 = 20$, $f_0 = 12$, $f_2 = 11$, $h = 10$

Formula: $$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

$$= 40 + \frac{20 - 12}{2(20) - 12 - 11} \times 10 = 40 + \frac{8}{40 - 23} \times 10$$

$$= 40 + \frac{8}{17} \times 10 = 40 + \frac{80}{17} = 40 + 4.7 = 44.7 \text{ (approx.)}$$

Answer: The mode of the data is 44.7 cars (approximately).

Given: Frequency distribution of monthly electricity consumption of 68 consumers.

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FINDING MEDIAN:

$n = 68$, so $\frac{n}{2} = 34$.

Cumulative Frequency Table:

| Class | $f_i$ | 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 |

The cumulative frequency just exceeding 34 is 42, so the median class = 125 – 145.

$l = 125$, $cf = 22$, $f = 20$, $h = 20$

$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 125 + \frac{34 - 22}{20} \times 20 = 125 + \frac{12}{20} \times 20 = 125 + 12 = 137$$

Median = 137 units

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FINDING MODE:

Highest frequency = 20, Modal class = 125 – 145.

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

$$\text{Mode} = 125 + \frac{20 - 13}{2(20) - 13 - 14} \times 20 = 125 + \frac{7}{40 - 27} \times 20 = 125 + \frac{7}{13} \times 20$$

$$= 125 + \frac{140}{13} = 125 + 10.77 = 135.77 \approx 135.8 \text{ units}$$

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FINDING MEAN:

Let assumed mean $a = 135$, $h = 20$.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i-135}{20}$ | $f_i u_i$ |
|---|---|---|---|---|
| 65 – 85 | 4 | 75 | –3 | –12 |
| 85 – 105 | 5 | 95 | –2 | –10 |
| 105 – 125 | 13 | 115 | –1 | –13 |
| 125 – 145 | 20 | 135 | 0 | 0 |
| 145 – 165 | 14 | 155 | 1 | 14 |
| 165 – 185 | 8 | 175 | 2 | 16 |
| 185 – 205 | 4 | 195 | 3 | 12 |
| Total | 68 | | | 7 |

$$\bar{x} = 135 + \frac{7}{68} \times 20 = 135 + \frac{140}{68} = 135 + 2.06 = 137.06 \approx 137.05 \text{ units}$$

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Comparison:
- Mean ≈ 137.05 units
- Median = 137 units
- Mode ≈ 135.8 units

All three values are very close to each other, indicating that the distribution is nearly symmetrical. The maximum number of consumers use electricity in the range 125–145 units.

Given: Median = 28.5, Total frequency = 60.

Step 1: Write the equation from total frequency.
$$5 + x + 20 + 15 + y + 5 = 60$$
$$45 + x + y = 60$$
$$x + y = 15 \quad \cdots (1)$$

Step 2: Since median = 28.5, the median class is 20 – 30.

$\frac{n}{2} = 30$

$l = 20$, $cf = 5 + x$, $f = 20$, $h = 10$

Formula: $$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$

$$28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10$$

$$28.5 - 20 = \frac{25 - x}{20} \times 10$$

$$8.5 = \frac{25 - x}{2}$$

$$17 = 25 - x$$

$$x = 8$$

Step 3: Substitute in equation (1):
$$8 + y = 15 \Rightarrow y = 7$$

Answer: $x = \mathbf{8}$ and $y = \mathbf{7}$.

Given: Cumulative frequency distribution of ages of 100 policy holders.

Step 1: Convert to class-wise frequency distribution.

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

Step 2: $n = 100$, $\frac{n}{2} = 50$.

The cumulative frequency just exceeding 50 is 78, so the median class = 35 – 40.

$l = 35$, $cf = 45$, $f = 33$, $h = 5$

Formula: $$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$

$$= 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{5}{33} \times 5 = 35 + \frac{25}{33}$$

$$= 35 + 0.76 = 35.76 \text{ years (approx.)}$$

Answer: The median age of the policy holders is 35.76 years (approximately).

Given: The data is in inclusive form. Convert to exclusive (continuous) form by subtracting 0.5 from lower limit and adding 0.5 to upper limit.

Step 1: Prepare the cumulative frequency table.

| Class (continuous) | $f_i$ | 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 |

Step 2: $n = 40$, $\frac{n}{2} = 20$.

The cumulative frequency just exceeding 20 is 29, so the median class = 144.5 – 153.5.

$l = 144.5$, $cf = 17$, $f = 12$, $h = 9$

Formula: $$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$

$$= 144.5 + \frac{20 - 17}{12} \times 9 = 144.5 + \frac{3}{12} \times 9 = 144.5 + \frac{27}{12}$$

$$= 144.5 + 2.25 = 146.75 \text{ mm}$$

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

Given: Frequency distribution of lifetimes of 400 neon lamps.

Step 1: Prepare the cumulative frequency table.

| Class | $f_i$ | 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 |

Step 2: $n = 400$, $\frac{n}{2} = 200$.

The cumulative frequency just exceeding 200 is 216, so the median class = 3000 – 3500.

$l = 3000$, $cf = 130$, $f = 86$, $h = 500$

Formula: $$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$

$$= 3000 + \frac{200 - 130}{86} \times 500 = 3000 + \frac{70}{86} \times 500$$

$$= 3000 + \frac{35000}{86} = 3000 + 406.98 = 3406.98 \text{ hours (approx.)}$$

Answer: The median lifetime of a lamp is 3406.98 hours (approximately).

Given: Frequency distribution of number of letters in 100 surnames.

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FINDING MEDIAN:

| Class | $f_i$ | 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 |

$n = 100$, $\frac{n}{2} = 50$.

The cumulative frequency just exceeding 50 is 76, so median class = 7 – 10.

$l = 7$, $cf = 36$, $f = 40$, $h = 3$

$$\text{Median} = 7 + \frac{50 - 36}{40} \times 3 = 7 + \frac{14}{40} \times 3 = 7 + \frac{42}{40} = 7 + 1.05 = 8.05$$

Median = 8.05 letters

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FINDING MEAN:

Let assumed mean $a = 8.5$, $h = 3$.

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i - 8.5}{3}$ | $f_i u_i$ |
|---|---|---|---|---|
| 1 – 4 | 6 | 2.5 | –2 | –12 |
| 4 – 7 | 30 | 5.5 | –1 | –30 |
| 7 – 10 | 40 | 8.5 | 0 | 0 |
| 10 – 13 | 16 | 11.5 | 1 | 16 |
| 13 – 16 | 4 | 14.5 | 2 | 8 |
| 16 – 19 | 4 | 17.5 | 3 | 12 |
| Total | 100 | | | –6 |

$$\bar{x} = 8.5 + \frac{-6}{100} \times 3 = 8.5 - 0.18 = 8.32$$

Mean = 8.32 letters

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FINDING MODE:

Highest frequency = 40, Modal class = 7 – 10.

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

$$\text{Mode} = 7 + \frac{40 - 30}{2(40) - 30 - 16} \times 3 = 7 + \frac{10}{80 - 46} \times 3 = 7 + \frac{10}{34} \times 3$$

$$= 7 + \frac{30}{34} = 7 + 0.88 = 7.88$$

Mode ≈ 7.88 letters

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Answer: Median = 8.05, Mean = 8.32, Mode ≈ 7.88 letters.

Given: Frequency distribution of weights of 30 students.

Step 1: Prepare the cumulative frequency table.

| Class | $f_i$ | Cumulative Frequency (cf) |
|---|---|---|
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 5 |
| 50 – 55 | 8 | 13 |
| 55 – 60 | 6 | 19 |
| 60 – 65 | 6 | 25 |
| 65 – 70 | 3 | 28 |
| 70 – 75 | 2 | 30 |

Step 2: $n = 30$, $\frac{n}{2} = 15$.

The cumulative frequency just exceeding 15 is 19, so the median class = 55 – 60.

$l = 55$, $cf = 13$, $f = 6$, $h = 5$

Formula: $$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h$$

$$= 55 + \frac{15 - 13}{6} \times 5 = 55 + \frac{2}{6} \times 5 = 55 + \frac{10}{6} = 55 + 1.67 = 56.67 \text{ kg (approx.)}$$

Answer: The median weight of the students is 56.67 kg (approximately).