Statistics — Class 10 Maths practice questions

Question 1

A district hospital in Pune conducted a health camp to study the age distribution of patients visiting the outpatient department over a period of one month. Dr. Meena Kulkarni, the chief medical officer, recorded the ages of 80 patients and organised the data into the following frequency distribution table:

| Age (in years) | Number of Patients |
|---|---|
| 10 – 20 | 6 |
| 20 – 30 | 11 |
| 30 – 40 | 21 |
| 40 – 50 | 23 |
| 50 – 60 | 14 |
| 60 – 70 | 5 |

This data was to be analysed to understand the age profile of patients and assist in better allocation of medical resources across different age groups.

Accepted Answer

**(i)**

The modal class is the class with the highest frequency.

From the table, the frequencies are 6, 11, 21, 23, 14, 5. The highest frequency is 23, corresponding to the class **40–50**.

∴ Modal class = **40–50**

---

**(ii)**

Total number of patients, n = 80 ⟹ n/2 = 40.

Cumulative frequencies:

| Age (in years) | Frequency | Cumulative Frequency |
|---|---|---|
| 10 – 20 | 6 | 6 |
| 20 – 30 | 11 | 17 |
| 30 – 40 | 21 | 38 |
| 40 – 50 | 23 | 61 |
| 50 – 60 | 14 | 75 |
| 60 – 70 | 5 | 80 |

The cumulative frequency first exceeds n/2 = 40 in the class **40–50**.

∴ Median class = **40–50**

---

**(iii)**

Using the Direct Method, Mean = Σfᵢxᵢ / Σfᵢ.

| Age (in years) | Midpoint xᵢ | Frequency fᵢ | fᵢxᵢ |
|---|---|---|---|
| 10 – 20 | 15 | 6 | 90 |
| 20 – 30 | 25 | 11 | 275 |
| 30 – 40 | 35 | 21 | 735 |
| 40 – 50 | 45 | 23 | 1035 |
| 50 – 60 | 55 | 14 | 770 |
| 60 – 70 | 65 | 5 | 325 |
| **Total** | | **Σfᵢ = 80** | **Σfᵢxᵢ = 3230** |

Mean = Σfᵢxᵢ / Σfᵢ = 3230 / 80

∴ Mean age = **40.375 years**

Question 2

A community health centre in Bhopal organised a free medical screening camp for residents of a nearby locality. Dr. Meena Sharma, the camp coordinator, recorded the ages of 80 patients who attended the camp and grouped the data into class intervals as shown in the table below.

| Age (in years) | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 |
|----------------|-------|-------|-------|-------|-------|-------|
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |

Accepted Answer

**(i) Modal Class**

The modal class is the class interval with the highest frequency.

From the table, the frequencies are: 6, 11, 21, 23, 14, 5.

The highest frequency is **23**, corresponding to the class interval **40–50**.

∴ The modal class is **40–50**.

---

**(ii) Median Class**

n = Σfᵢ = 80 ⟹ n/2 = 40

Cumulative frequency table:

| Age (in years) | Frequency (fᵢ) | Cumulative Frequency (cf) |
|----------------|----------------|---------------------------|
| 10–20 | 6 | 6 |
| 20–30 | 11 | 17 |
| 30–40 | 21 | 38 |
| 40–50 | 23 | 61 |
| 50–60 | 14 | 75 |
| 60–70 | 5 | 80 |

The cumulative frequency first exceeds n/2 = 40 in the class **40–50** (cf rises from 38 to 61).

∴ The median class is **40–50**.

---

**(iii) Mean Age**

Using the Direct Method: Mean = Σfᵢxᵢ / Σfᵢ

| Age (in years) | fᵢ | xᵢ (mid-value) | fᵢxᵢ |
|----------------|-----|----------------|-------|
| 10–20 | 6 | 15 | 90 |
| 20–30 | 11 | 25 | 275 |
| 30–40 | 21 | 35 | 735 |
| 40–50 | 23 | 45 | 1035 |
| 50–60 | 14 | 55 | 770 |
| 60–70 | 5 | 65 | 325 |
| **Total** | **80** | | **3230** |

Mean = Σfᵢxᵢ / Σfᵢ = 3230 / 80

∴ Mean age = **40.375 years**

Question 3

A municipal library in Pune recorded the number of books borrowed by its members during a month. The data for 60 members is given below:

| Number of books borrowed | 0–5 | 5–10 | 10–15 | 15–20 | 20–25 |
|---|---|---|---|---|---|
| Number of members | 6 | 10 | 18 | 14 | 12 |

The librarian wants to analyse borrowing habits by constructing ogives and estimating the median number of books borrowed.

Accepted Answer

**(i)**

Cumulative frequency table ('less than' type):

| Upper Boundary | Cumulative Frequency |
|---|---|
| Less than 5 | 6 |
| Less than 10 | 6 + 10 = 16 |
| Less than 15 | 16 + 18 = **34** |

∴ The cumulative frequency corresponding to the class 10–15 (i.e., less than 15) is **34**.

---

**(ii)**

'More than 10' cumulative frequency = Total members − (members in 0–5 and 5–10)

⟹ = 60 − 6 − 10 = **44**

∴ The coordinates of the point on the 'more than' ogive for class boundary 10 are **(10, 44)**.

---

**(iii)**

**'Less than' cumulative frequency table:**

| Upper Boundary | Cumulative Frequency (cf) |
|---|---|
| Less than 5 | 6 |
| Less than 10 | 16 |
| Less than 15 | 34 |
| Less than 20 | 48 |
| Less than 25 | 60 |

n = Σf = 60 ∴ n/2 = 30

The cumulative frequency first exceeds 30 in the class **10–15**.

∴ Median class = 10–15, where l = 10, cf = 16, f = 18, h = 5.

Applying the Median formula:

Median = l + [(n/2 − cf) / f] × h

⟹ Median = 10 + [(30 − 16) / 18] × 5

⟹ Median = 10 + [14/18] × 5

⟹ Median = 10 + 70/18

⟹ Median = 10 + 35/9

∴ Median = 125/9 ≈ **13.9 books**

Question 4

A municipal corporation recorded the daily water consumption (in litres) of 60 households in a newly developed colony. The data collected was grouped into class intervals as follows: 0–50: 4 households, 50–100: 8 households, 100–150: 14 households, 150–200: 18 households, 200–250: 10 households, 250–300: 6 households. The data is being analysed to understand consumption patterns and plan water supply accordingly.

Accepted Answer

**(i)**

Cumulative frequency up to the class 150–200:

cf = 4 + 8 + 14 + 18 = **44**

∴ The cumulative frequency up to the class 150–200 is **44**.

---

**(ii)**

Total number of households, n = Σfᵢ = 4 + 8 + 14 + 18 + 10 + 6 = 60

⟹ n/2 = 60/2 = **30**

∴ The value of n/2 is **30**.

---

**(iii)**

**'Less than' Ogive — points to be plotted:**

| Upper Class Boundary | Cumulative Frequency |
|---|---|
| 50 | 4 |
| 100 | 12 |
| 150 | 26 |
| 200 | 44 |
| 250 | 54 |
| 300 | 60 |

Plot the points (50, 4), (100, 12), (150, 26), (200, 44), (250, 54), (300, 60) on a graph with upper class boundaries on the x-axis and cumulative frequency on the y-axis. Join them with a smooth freehand curve to obtain the 'less than' ogive.

**Reading the Median from the Ogive:**

Draw a horizontal line from n/2 = 30 on the cumulative frequency axis to meet the ogive, then draw a vertical line from that point to the x-axis. The x-value obtained is the median.

Since cf = 26 at x = 150 and cf = 44 at x = 200, the point corresponding to cf = 30 lies in the interval 150–200, giving median ≈ 161.1 litres.

**Verification using the Median Formula:**

Median class = 150–200  (∵ cumulative frequency first exceeds n/2 = 30 in this class)

Median = l + [(n/2 − cf) / f] × h

where l = 150, cf = 26, f = 18, h = 50

⟹ Median = 150 + [(30 − 26) / 18] × 50

⟹ Median = 150 + [4/18] × 50

⟹ Median = 150 + 200/18

⟹ Median = 150 + 11.1̄

∴ **Median daily water consumption ≈ 161.1 litres**

Question 5

A school in Jaipur conducted a survey to study the screen time (in hours per week) of 60 students from Class 10. The data collected was organised into the following frequency distribution table: Screen time (hrs/week): 0–5, 5–10, 10–15, 15–20, 20–25, 25–30; Number of students: 4, 8, 14, 20, 10, 4. The school counsellor decided to draw ogives to analyse the data and find the median screen time of the students.

Accepted Answer

**(i)**

Cumulative frequency table (less than type):

| Screen Time (hrs/week) | Frequency | Cumulative Frequency |
|------------------------|-----------|----------------------|
| Less than 5 | 4 | 4 |
| Less than 10 | 8 | 12 |
| Less than 15 | 14 | 26 |
| Less than 20 | 20 | 46 |

∴ The cumulative frequency for 'less than 20' is **46**.

---

**(ii)**

For the less than ogive, each point is plotted as (upper class boundary, cumulative frequency up to that boundary).

For class 15–20: upper boundary = 20, cumulative frequency = 4 + 8 + 14 + 20 = 46.

∴ The point to be plotted on the less than ogive for class 15–20 is **(20, 46)**.

---

**(iii)**

Complete cumulative frequency table:

| Screen Time (hrs/week) | Frequency (f) | Cumulative Frequency (cf) |
|------------------------|---------------|--------------------------|
| 0–5 | 4 | 4 |
| 5–10 | 8 | 12 |
| 10–15 | 14 | 26 |
| **15–20** | **20** | **46** |
| 20–25 | 10 | 56 |
| 25–30 | 4 | 60 |

Here, n = Σfᵢ = 60 ⟹ n/2 = 30.

The median class is the class whose cumulative frequency first exceeds n/2 = 30.

cf just before class 15–20 = 26 < 30, and cf up to 15–20 = 46 ≥ 30.

∴ Median class = **15–20**.

Using the Median formula:

Median = l + [(n/2 − cf) / f] × h

where l = 15, cf = 26, f = 20, h = 5.

⟹ Median = 15 + [(30 − 26) / 20] × 5

⟹ Median = 15 + [4/20] × 5

⟹ Median = 15 + 1

∴ **Median screen time = 16 hours per week.**

*(Ogive verification: On the less than ogive, a horizontal line drawn from 30 on the y-axis meets the curve in the class 15–20; the perpendicular dropped from that point to the x-axis touches at x = 16, confirming the median.)*

Question 6

A school health committee conducted a fitness survey of 60 students in Class 10. The time (in minutes) each student spends on physical activity per day was recorded and grouped as follows:

| Time (in minutes) | Number of Students |
|-------------------|--------------------|
| 10 – 20           | 4                  |
| 20 – 30           | 8                  |
| 30 – 40           | 14                 |
| 40 – 50           | 18                 |
| 50 – 60           | 10                 |
| 60 – 70           | 6                  |

The committee wishes to analyse the data to understand the central tendency of students' physical activity habits.

(i) Find the modal class and hence calculate the Mode of the data.  [1]
(ii) Using the assumed mean method with a = 45, calculate the Mean time spent on physical activity.  [1]
(iii) Find the Median time spent on physical activity. Hence verify the empirical relationship: Mode = 3 × Median − 2 × Mean.  [2]

Accepted Answer

(i) The class with the highest frequency is 40–50, with frequency 18.
∴ Modal class = 40–50

Using the Mode formula:
Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h

where l = 40, f₁ = 18, f₀ = 14, f₂ = 10, h = 10

⟹ Mode = 40 + [(18 − 14) / (2×18 − 14 − 10)] × 10
⟹ Mode = 40 + [4 / (36 − 24)] × 10
⟹ Mode = 40 + (4/12) × 10
⟹ Mode = 40 + 40/12
⟹ Mode = 40 + 3.33
∴ Mode = 43.33 minutes

---

(ii) Let assumed mean a = 45, h = 10.

Construct the table for assumed mean method:

| Class   | fᵢ | xᵢ (mid-value) | dᵢ = xᵢ − a | fᵢdᵢ |
|---------|----|----------------|-------------|------|
| 10–20   | 4  | 15             | −30         | −120 |
| 20–30   | 8  | 25             | −20         | −160 |
| 30–40   | 14 | 35             | −10         | −140 |
| 40–50   | 18 | 45             |   0         |    0 |
| 50–60   | 10 | 55             |  10         |  100 |
| 60–70   | 6  | 65             |  20         |  120 |
| **Total** | **60** |            |             | **−200** |

n = Σfᵢ = 60,  Σfᵢdᵢ = −200

Mean = a + (Σfᵢdᵢ / Σfᵢ)
⟹ Mean = 45 + (−200 / 60)
⟹ Mean = 45 − 3.33
∴ Mean = 41.67 minutes

---

(iii) Construct the cumulative frequency table:

| Class   | fᵢ | Cumulative Frequency (cf) |
|---------|----|---------------------------|
| 10–20   | 4  | 4                         |
| 20–30   | 8  | 12                        |
| 30–40   | 14 | 26                        |
| 40–50   | 18 | 44                        |
| 50–60   | 10 | 54                        |
| 60–70   | 6  | 60                        |

n = 60  ⟹  n/2 = 30

The cumulative frequency first exceeding 30 is 44, corresponding to class 40–50.
∴ Median class = 40–50

Using the Median formula:
Median = l + [(n/2 − cf) / f] × h

where l = 40, cf = 26, f = 18, h = 10

⟹ Median = 40 + [(30 − 26) / 18] × 10
⟹ Median = 40 + (4/18) × 10
⟹ Median = 40 + 40/18
⟹ Median = 40 + 2.22
∴ Median = 42.22 minutes

Verification of Empirical Relationship: Mode = 3 × Median − 2 × Mean

RHS = 3 × 42.22 − 2 × 41.67
⟹ RHS = 126.66 − 83.34
⟹ RHS = 43.32 ≈ 43.33

LHS = Mode = 43.33

∴ LHS ≈ RHS

Hence, the empirical relationship Mode = 3 × Median − 2 × Mean is verified.

Question 7

A municipal corporation conducted a survey on the monthly electricity consumption (in units) of 60 households in a residential colony. The data collected was organized into the following frequency distribution table: Monthly Consumption (in units): 0–50, 50–100, 100–150, 150–200, 200–250, 250–300; Number of Households: 4, 8, 14, 18, 10, 6. The corporation wants to analyze this data using ogives to determine the median consumption and make decisions about electricity supply.

Accepted Answer

**(i)**

| Class | Frequency | Cumulative Frequency (less than) |
|-------|-----------|----------------------------------|
| 0–50 | 4 | 4 |
| 50–100 | 8 | 12 |
| 100–150 | 14 | 26 |
| 150–200 | 18 | 44 |
| 200–250 | 10 | 54 |
| 250–300 | 6 | 60 |

The cumulative frequency corresponding to the upper boundary of class 150–200 is 26 + 18 = 44.

∴ The cumulative frequency of the class 150–200 is **44**.

---

**(ii)**

For a 'more than' ogive, the cumulative frequency at a lower class boundary = Total frequency − cumulative frequency of all classes below that boundary.

Total n = 60.

Cumulative frequency of all classes below 100 = 4 + 8 = 12.

⟹ 'More than' cumulative frequency at lower boundary 100 = 60 − 12 = **48**.

∴ The 'more than' cumulative frequency at 100 units is **48**.

---

**(iii)**

Cumulative frequency table (less than type):

| Class | Frequency (f) | Cumulative Frequency (cf) |
|-------|--------------|--------------------------|
| 0–50 | 4 | 4 |
| 50–100 | 8 | 12 |
| 100–150 | 14 | 26 |
| 150–200 | 18 | 44 |
| 200–250 | 10 | 54 |
| 250–300 | 6 | 60 |

n = 60, ∴ n/2 = 30.

The cumulative frequency first exceeding 30 is 44, which belongs to the class **150–200**.

∴ Median class = 150–200, where l = 150, cf = 26, f = 18, h = 50.

Applying the Median formula:

Median = l + [(n/2 − cf) / f] × h

⟹ Median = 150 + [(30 − 26) / 18] × 50

⟹ Median = 150 + [4/18] × 50

⟹ Median = 150 + 200/18

⟹ Median = 150 + 11.11

∴ The median monthly electricity consumption is approximately **161.11 units**.

Question 8

A hospital recorded the ages (in years) of 80 patients admitted to its emergency ward during a particular month. The data was grouped into class intervals as shown below:

| Age (in years) | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 |
|---|---|---|---|---|---|---|
| Number of Patients | 6 | 10 | 18 | 22 | 16 | 8 |

The hospital administration wants to analyse the age distribution using cumulative frequency methods to determine the median age and draw ogives for better understanding.

Accepted Answer

**(i)**

Cumulative frequency table ('less than' type):

| Age (in years) | Frequency | Cumulative Frequency |
|---|---|---|
| Less than 20 | 6 | 6 |
| Less than 30 | 10 | 16 |
| Less than 40 | 18 | 34 |
| Less than 50 | 22 | 56 |
| Less than 60 | 16 | 72 |
| Less than 70 | 8 | 80 |

Cumulative frequency up to 40 (i.e., up to the end of class 30–40):

= 6 + 10 + 18 ⟹ **∴ Cumulative frequency = 34**

---

**(ii)**

For the 'more than' ogive, the cumulative frequency corresponding to 'more than 40' counts all patients in classes 40–50, 50–60, and 60–70.

= 22 + 16 + 8 ⟹ **∴ Cumulative frequency for 'more than 40' = 46**

---

**(iii)**

**Cumulative Frequency Table:**

| Age (in years) | Number of Patients (f) | Cumulative Frequency (cf) |
|---|---|---|
| 10–20 | 6 | 6 |
| 20–30 | 10 | 16 |
| 30–40 | 18 | 34 |
| 40–50 | 22 | 56 |
| 50–60 | 16 | 72 |
| 60–70 | 8 | 80 |

Here, n = Σfᵢ = 80 ⟹ n/2 = 40.

The median class is the class whose cumulative frequency first exceeds n/2 = 40.

cf up to 30–40 = 34 < 40; cf up to 40–50 = 56 ≥ 40.

∴ **Median class = 40–50.**

Applying the Median formula:

$$	ext{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}ight) 	imes h$$

where l = 40, n/2 = 40, cf = 34, f = 22, h = 10.

$$	ext{Median} = 40 + \left(\frac{40 - 34}{22}ight) 	imes 10$$

$$= 40 + \frac{6 	imes 10}{22} = 40 + \frac{60}{22}$$

$$= 40 + 2.73$$

**∴ Median age ≈ 42.73 years.**

Question 9

Consider the following frequency distribution of the daily wages (in ₹) of 60 workers in a factory:

Daily Wages (₹): 100–120, 120–140, 140–160, 160–180, 180–200
Number of Workers: 8, 12, 20, 14, 6

The sum of the lower limits of the median class and the modal class is:

Accepted Answer

Option (A) is correct.

**Explanation:** Using Median class: n = 60 ⟹ n/2 = 30. Cumulative frequencies: 100–120: 8; 120–140: 20; 140–160: 40; 160–180: 54; 180–200: 60. The cumulative frequency first exceeds 30 in the class 140–160 (cf = 40, previous cf = 20), ∴ median class = 140–160, lower limit = 140. Modal class has the highest frequency = 20 (class 140–160), ∴ modal class = 140–160, lower limit = 140. ∴ Sum = 140 + 140 = **280**.

Question 10

The following frequency distribution shows the number of patients admitted per day in a hospital over 90 days:

| Patients per Day | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Number of Days | 8 | 17 | 29 | 24 | 12 |

The median of the above data is:

Accepted Answer

Option (B) is correct.

**Explanation:** n = Σfᵢ = 8 + 17 + 29 + 24 + 12 = 90, so n/2 = 45. Cumulative frequencies: 0–10: 8; 0–20: 25; 0–30: 54. Since cf first exceeds 45 in the class 20–30, the median class is 20–30, with l = 20, cf = 25, f = 29, h = 10. Median = l + [(n/2 − cf)/f] × h = 20 + [(45 − 25)/29] × 10 = 20 + 200/29 ≈ 20 + 6.90 = **26.90**.

Question 11

Consider the following frequency distribution of the number of hours of screen time per day recorded for 80 students:

| Screen Time (hours) | 0–2 | 2–4 | 4–6 | 6–8 | 8–10 |
|---|---|---|---|---|---|
| Number of Students | 5 | 13 | 24 | 26 | 12 |

The lower limit of the median class is:

Accepted Answer

Option (A) is correct.

**Explanation:** n = Σfᵢ = 80, so n/2 = 40. The cumulative frequencies are: 0–2: 5; 0–4: 18; 0–6: 42; 0–8: 68; 0–10: 80. The median class is the class whose cumulative frequency first exceeds n/2 = 40, which is the class 4–6 (cf = 42). ∴ the lower limit of the median class is **4**.

Question 12

The following frequency distribution shows the number of books borrowed from a library each week over 60 weeks:

| Books Borrowed | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|
| Number of Weeks | 5 | 8 | 16 | 22 | 9 |

The sum of the lower limits of the median class and the modal class is:

Accepted Answer

Option (C) is correct.

Explanation: n = 60, so n/2 = 30. The cumulative frequencies are: 10–20 → 5; 20–30 → 13; 30–40 → 29; 40–50 → 51. The cumulative frequency first exceeding 30 is 51, so the median class is 40–50 (lower limit = 40). The class with the highest frequency is 40–50 (f = 22), so the modal class is also 40–50 (lower limit = 40). ∴ Sum of lower limits = 40 + 40 = 80.

Question 13

Consider the following frequency distribution of the number of pages read per day by 50 students:

| Pages per Day | 0–8 | 8–16 | 16–24 | 24–32 | 32–40 |
|---|---|---|---|---|---|
| Number of Students | 4 | 9 | 18 | 13 | 6 |

The sum of the lower limits of the median class and the modal class is:

Accepted Answer

Option (B) is correct.

**Explanation:** For median class, n/2 = 50/2 = 25. Cumulative frequencies: 0–8: 4; 0–16: 13; 0–24: 31. Since cf first exceeds 25 in the class 16–24, the median class is **16–24** (lower limit = 16). For modal class, the highest frequency is 18, corresponding to class 16–24, so the modal class is also **16–24** (lower limit = 16). ∴ Sum of lower limits = 16 + 16 = **32**.

Question 14

Consider the following frequency distribution of the number of books read by 60 students in a year:

| Number of Books | 0–4 | 4–8 | 8–12 | 12–16 | 16–20 |
|---|---|---|---|---|---|
| Number of Students | 7 | 12 | 20 | 14 | 7 |

The sum of the lower limits of the median class and the modal class is:

Accepted Answer

Option (B) is correct.

**Explanation:** For median class, n = 60 ⟹ n/2 = 30. Cumulative frequencies: 0–4: 7; 0–8: 19; 0–12: 39. Since cf first exceeds 30 in the class 8–12 (cf goes from 19 to 39), the median class is **8–12**, lower limit = 8. For modal class, the highest frequency is 20, corresponding to class **8–12**, lower limit = 8. ∴ Sum = 8 + 8 = **16**.

Question 15

Consider the following frequency distribution of the number of books read by 60 students in a year:

| Number of Books | 0–6 | 6–12 | 12–18 | 18–24 | 24–30 |
|---|---|---|---|---|---|
| Number of Students | 5 | 12 | 20 | 16 | 7 |

The sum of the lower limits of the median class and the modal class is:

Accepted Answer

Option (C) is correct.

**Explanation:** For n = 60, n/2 = 30. The cumulative frequencies are: 0–6: 5, 6–12: 17, 12–18: 37, 18–24: 53, 24–30: 60. The cumulative frequency first exceeds 30 in the class 12–18, so the median class is **12–18** with lower limit **12**. The highest frequency is 20, corresponding to class 12–18, so the modal class is also **12–18** with lower limit **12**. ∴ Sum of lower limits = 12 + 12 = **24**.

Question 16

If the mean and mode of a frequency distribution are 45 and 33 respectively, then its median is:

Accepted Answer

Option (B) is correct.

Explanation: By the empirical relationship Mode = 3·Median − 2·Mean, substituting Mode = 33 and Mean = 45 ⟹ 33 = 3·Median − 2×45 ⟹ 3·Median = 33 + 90 = 123 ⟹ Median = 41.

Question 17

For a frequency distribution, if the mean is 36 and the mode is 24, then the median of the distribution is:

Accepted Answer

Option (B) is correct.

Explanation: By the empirical relationship Mode = 3 × Median − 2 × Mean, substituting Mode = 24 and Mean = 36: 24 = 3 × Median − 2 × 36 ⟹ 3 × Median = 24 + 72 = 96 ⟹ Median = 32.

Question 18

The median class of the following frequency distribution of the number of customers visiting a store over 50 days is:

| Customers | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | 100–120 |
|---|---|---|---|---|---|---|
| Number of Days | 4 | 6 | 14 | 16 | 8 | 2 |

Accepted Answer

Option (B) is correct.

Explanation: The median class is the class whose cumulative frequency first exceeds n/2. Here n = Σfᵢ = 50, so n/2 = 25. The cumulative frequencies are: 0–20: 4; 0–40: 10; 0–60: 24; 0–80: 40. Since the cumulative frequency 40 (for class 60–80) is the first to exceed 25, the median class is **60–80**.

Question 19

The following frequency distribution shows the number of hours spent on studying per week by 40 students:

| Study Hours | 0–5 | 5–10 | 10–15 | 15–20 | 20–25 |
|---|---|---|---|---|---|
| Number of Students | 3 | 7 | 13 | 12 | 5 |

The lower limit of the median class is:

Accepted Answer

Option (B) is correct.

**Explanation:** For a grouped frequency distribution, the median class is the class whose cumulative frequency first exceeds n/2. Here n = 40, so n/2 = 20. The cumulative frequencies are: 0–5 → 3; 5–10 → 10; 10–15 → 23. Since 23 is the first cumulative frequency to exceed 20, the median class is 10–15, whose lower limit is **10**.

Question 20

For a frequency distribution, if mean = 42 and mode = 36, then the median of the distribution is:

Accepted Answer

Option (B) is correct.

Explanation: By the empirical relationship Mode = 3 Median − 2 Mean, substituting Mode = 36 and Mean = 42: 36 = 3 × Median − 2 × 42 ⟹ 3 × Median = 36 + 84 = 120 ⟹ Median = 40.

Statistics: Class 10 Maths Practice Questions

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