The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above.

Mode = 36.8 years. Mean = 35.375 years. The maximum number of patients admitted are of age 36.8 years, while the average age is 35.375 years.

NCERT Solutions Class 10 Maths Chapter 13: Statistics

Q: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetimes of the components.

The modal lifetime of the components is 65.625 hours.

Q: The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families.

The modal monthly expenditure of the families is ₹ 1847.83.

Q: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.

The mode of the data is 4608.7 runs.

Q: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

The mode of the data is 44.7 cars.

Detailed solutions for Exercise 13.2 (Mode of Grouped Data).

Ex 13.1 Ex 13.2 Ex 13.3

#### **Q1. The following table shows the ages of the patients admitted in a hospital during a year:** | Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of patients** | 6 | 11 | 21 | 23 | 14 | 5 | **Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.** #### **A1. Solution:** **Finding Mode:** Maximum frequency is 23. So, the modal class is **35 - 45**. $l = 35$ (lower limit of modal class) $h = 10$ (class size) $f_1 = 23$ (frequency of modal class) $f_0 = 21$ (frequency of class preceding modal class) $f_2 = 14$ (frequency of class succeeding modal class) Mode = $l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$ $= 35 + \left( \frac{23 - 21}{2(23) - 21 - 14} \right) \times 10$ $= 35 + \left( \frac{2}{46 - 35} \right) \times 10$ $= 35 + \frac{20}{11} = 35 + 1.818...$ $= 36.8$ years (approx). **Finding Mean:** Using Assumed Mean Method ($a = 30$, $h = 10$). $\sum f_i = 80$, $\sum f_i u_i = 43$. Mean $\bar{x} = a + (\frac{\sum f_i u_i}{\sum f_i}) \times h$ $= 30 + (\frac{43}{80}) \times 10 = 30 + 5.375 = 35.375$ years. **Interpretation:** The mode is 36.8 years, and the mean is 35.375 years. This implies that the maximum number of patients admitted in the hospital are of age 36.8 years, while on average, the age of a patient admitted is 35.375 years. #### **Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:** | Lifetimes (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | **Frequency** | 10 | 35 | 52 | 61 | 38 | 29 | **Determine the modal lifetimes of the components.** #### **A2. Solution:** Maximum frequency is 61. So, the modal class is **60 - 80**. $l = 60, h = 20, f_1 = 61, f_0 = 52, f_2 = 38$. Mode = $l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$ $= 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20$ $= 60 + \left( \frac{9}{122 - 90} \right) \times 20$ $= 60 + \left( \frac{9}{32} \right) \times 20$ $= 60 + \frac{180}{32} = 60 + 5.625$ $= 65.625$ hours. #### **Q3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.** | Expenditure (in ₹) | 1000-1500 | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of families** | 24 | 40 | 33 | 28 | 30 | 22 | 16 | 7 | #### **A3. Solution:** **Finding Mode:** Maximum frequency is 40. Modal class is **1500 - 2000**. $l = 1500, h = 500, f_1 = 40, f_0 = 24, f_2 = 33$. Mode = $1500 + \left( \frac{40 - 24}{2(40) - 24 - 33} \right) \times 500$ $= 1500 + \left( \frac{16}{80 - 57} \right) \times 500$ $= 1500 + \frac{16}{23} \times 500$ $= 1500 + \frac{8000}{23} \approx 1500 + 347.83$ $= 1847.83$ ₹. **Finding Mean:** Using Step Deviation Method ($a = 2750, h = 500$). Mean $\bar{x} = 2662.5$ ₹. #### **Q4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.** | Number of students per teacher | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of states/U.T.** | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 | #### **A4. Solution:** **Finding Mode:** Maximum frequency is 10. Modal class is **30 - 35**. $l = 30, h = 5, f_1 = 10, f_0 = 9, f_2 = 3$. Mode = $30 + \left( \frac{10 - 9}{2(10) - 9 - 3} \right) \times 5$ $= 30 + \left( \frac{1}{20 - 12} \right) \times 5$ $= 30 + \frac{5}{8} = 30 + 0.625$ $= 30.6$ (approx). **Finding Mean:** Using Step Deviation Method ($a = 32.5, h = 5$). Mean $\bar{x} = 29.2$. **Interpretation:** Most states have a student-teacher ratio of 30.6, while the average ratio is 29.2. #### **Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.** | Runs scored | 3000-4000 | 4000-5000 | 5000-6000 | 6000-7000 | 7000-8000 | 8000-9000 | 9000-10000 | 10000-11000 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of batsmen** | 4 | 18 | 9 | 7 | 6 | 3 | 1 | 1 | **Find the mode of the data.** #### **A5. Solution:** Maximum frequency is 18. Modal class is **4000 - 5000**. $l = 4000, h = 1000, f_1 = 18, f_0 = 4, f_2 = 9$. Mode = $4000 + \left( \frac{18 - 4}{2(18) - 4 - 9} \right) \times 1000$ $= 4000 + \left( \frac{14}{36 - 13} \right) \times 1000$ $= 4000 + \frac{14000}{23} \approx 4000 + 608.7$ $= 4608.7$ runs. #### **Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.** | Number of cars | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Frequency** | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 | #### **A6. Solution:** Maximum frequency is 20. Modal class is **40 - 50**. $l = 40, h = 10, f_1 = 20, f_0 = 12, f_2 = 11$. Mode = $40 + \left( \frac{20 - 12}{2(20) - 12 - 11} \right) \times 10$ $= 40 + \left( \frac{8}{40 - 23} \right) \times 10$ $= 40 + \frac{80}{17} \approx 40 + 4.7$ $= 44.7$ cars.