The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

With N=400, the median class is 3000-3500. Using the formula Median = l + [(N/2 - cf)/f] × h, where l=3000, cf=130, f=86, and h=500, the median life time is calculated to be approximately 3406.98 hours.

100 surnames were randomly picked up from a local telephone directory. Determine the median number of letters in the surnames. Find the mean and modal size.

For the given data: The Median number of letters is 8.05. The Mean number of letters is 8.32. The Modal size of the surnames is 7.88.

NCERT Solutions Class 10 Maths Chapter 13: Statistics

Q: If the median of the distribution given below is 28.5, find the values of x and y. (Class interval: 0-10, 10-20... Total Frequency: 60)

Given the median is 28.5 and total frequency (N) is 60. By using the median formula and the equation x + y = 15 derived from the total frequency, we find that x = 8 and y = 7.

Q: A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

First, convert the 'Below' cumulative frequency distribution into a standard frequency distribution. Using N=100, the median class is found to be 35-40. Applying the median formula, the median age is calculated as 35.76 years.

Detailed solutions for Exercise 13.3 (Median of Grouped Data).

Ex 13.1 Ex 13.2 Ex 13.3

#### **Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.** | Monthly consumption (in units) | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of consumers** | 4 | 5 | 13 | 20 | 14 | 8 | 4 | #### **A1. Solution:** **1. Median:** $N = 68 \Rightarrow \frac{N}{2} = 34$. Cumulative Frequency (cf) just greater than 34 is 42, corresponding to class **125 - 145**. $l = 125, h = 20, f = 20, cf = 22$ (preceding class). Median = $l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ $= 125 + \left( \frac{34 - 22}{20} \right) \times 20$ $= 125 + 12 = 137$ units. **2. Mean:** Using Step Deviation Method ($a = 135, h = 20$). Mean $\bar{x} = 137.05$ units. **3. Mode:** Max frequency is 20. Modal class is **125 - 145**. $l = 125, f_1 = 20, f_0 = 13, f_2 = 14$. Mode = $125 + \left( \frac{20 - 13}{40 - 13 - 14} \right) \times 20$ $= 125 + \frac{140}{13} = 135.76$ units. **Comparison:** Median (137) > Mean (137.05) > Mode (135.76). The three measures are approximately the same. #### **Q2. If the median of the distribution given below is 28.5, find the values of $x$ and $y$.** | Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | **Total** | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Frequency** | 5 | $x$ | 20 | 15 | $y$ | 5 | **60** | #### **A2. Solution:** Given: $N = 60$, Median = 28.5. Since Median is 28.5, the median class is **20 - 30**. $l = 20, h = 10, f = 20$. Cumulative Frequency ($cf$) of preceding class = $5 + x$. 1. Equation from Total Frequency: $5 + x + 20 + 15 + y + 5 = 60$ $45 + x + y = 60 \Rightarrow x + y = 15$ ...(i) 2. Median Formula: Median = $l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ $28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10$ $8.5 = \frac{25 - x}{2}$ $17 = 25 - x \Rightarrow x = 8$. Substitute $x = 8$ in (i): $8 + y = 15 \Rightarrow y = 7$. **Values are $x = 8$ and $y = 7$.** #### **Q3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.** | Age (in years) | Below 20 | Below 25 | Below 30 | Below 35 | Below 40 | Below 45 | Below 50 | Below 55 | Below 60 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of policy holders** | 2 | 6 | 24 | 45 | 78 | 89 | 92 | 98 | 100 | #### **A3. Solution:** Convert cumulative frequency to simple frequency distribution: 15-20: 2 20-25: $6-2=4$ 25-30: $24-6=18$ 30-35: $45-24=21$ 35-40: $78-45=33$ 40-45: $89-78=11$ 45-50: $92-89=3$ 50-55: $98-92=6$ 55-60: $100-98=2$ $N = 100 \Rightarrow \frac{N}{2} = 50$. Class with cf > 50 is **35 - 40** (cf = 78). $l = 35, h = 5, f = 33, cf = 45$. Median = $35 + \left( \frac{50 - 45}{33} \right) \times 5$ $= 35 + \frac{25}{33} = 35 + 0.76$ $= 35.76$ years. #### **Q4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:** | Length (in mm) | 118-126 | 127-135 | 136-144 | 145-153 | 154-162 | 163-171 | 172-180 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of leaves** | 3 | 5 | 9 | 12 | 5 | 4 | 2 | **Find the median length of the leaves.** #### **A4. Solution:** The series is discontinuous. Convert to continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit. Classes: 117.5-126.5, 126.5-135.5, etc. $N = 40 \Rightarrow \frac{N}{2} = 20$. Cumulative frequencies: 3, 8, 17, 29, 34, 38, 40. Class with cf > 20 is **144.5 - 153.5** (cf = 29). $l = 144.5, h = 9, f = 12, cf = 17$. Median = $144.5 + \left( \frac{20 - 17}{12} \right) \times 9$ $= 144.5 + \frac{3}{12} \times 9$ $= 144.5 + 2.25 = 146.75$ mm. #### **Q5. The following table gives the distribution of the life time of 400 neon lamps:** | Life time (in hours) | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of lamps** | 14 | 56 | 60 | 86 | 74 | 62 | 48 | **Find the median life time of a lamp.** #### **A5. Solution:** First, we calculate the cumulative frequencies ($cf$). | Life time (in hours) | Frequency ($f$) | 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** | **$N = 400$** | | Here, $N = 400$, so $\frac{N}{2} = 200$. The cumulative frequency just greater than 200 is 216, which corresponds to the class interval **3000 - 3500**. Thus, the median class is **3000 - 3500**. Given: Lower limit ($l$) = 3000 Class size ($h$) = 500 Frequency of median class ($f$) = 86 Cumulative frequency of preceding class ($cf$) = 130 Median = $l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ $= 3000 + \left( \frac{200 - 130}{86} \right) \times 500$ $= 3000 + \left( \frac{70}{86} \right) \times 500$ $= 3000 + \frac{35000}{86}$ $= 3000 + 406.976...$ $\approx 3406.98$ hours. **The median life time of a lamp is 3406.98 hours.** #### **Q6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:** | Number of letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of surnames** | 6 | 30 | 40 | 16 | 4 | 4 | **Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.** #### **A6. Solution:** **1. Median:** $N = 100$, so $\frac{N}{2} = 50$. Cumulative frequencies: 1-4: 6 4-7: 36 7-10: 76 (This is $> 50$, so median class is **7 - 10**) 10-13: 92 ... Median Class: **7 - 10** $l = 7, h = 3, f = 40, cf = 36$. Median = $7 + \left( \frac{50 - 36}{40} \right) \times 3$ $= 7 + \frac{14}{40} \times 3$ $= 7 + \frac{42}{40} = 7 + 1.05 = 8.05$. **2. Mean:** Using Step Deviation Method ($a = 8.5, h = 3$). Class marks ($x_i$): 2.5, 5.5, 8.5, 11.5, 14.5, 17.5. Mean $\bar{x} = 8.32$. **3. Mode:** Maximum frequency is 40. Modal class is **7 - 10**. $l = 7, h = 3, f_1 = 40, f_0 = 30, f_2 = 16$. Mode = $7 + \left( \frac{40 - 30}{2(40) - 30 - 16} \right) \times 3$ $= 7 + \left( \frac{10}{80 - 46} \right) \times 3$ $= 7 + \frac{30}{34} = 7 + 0.88 = 7.88$. **Median = 8.05, Mean = 8.32, Mode = 7.88.** #### **Q7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.** | Weight (in kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Number of students** | 2 | 3 | 8 | 6 | 6 | 3 | 2 | #### **A7. Solution:** Calculate cumulative frequencies ($cf$): 40-45: 2 45-50: 5 50-55: 13 55-60: 19 60-65: 25 65-70: 28 70-75: 30 $N = 30$, so $\frac{N}{2} = 15$. Cumulative frequency just greater than 15 is 19, corresponding to class **55 - 60**. Median Class: **55 - 60**. $l = 55, h = 5, f = 6, cf = 13$. Median = $l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ $= 55 + \left( \frac{15 - 13}{6} \right) \times 5$ $= 55 + \frac{2}{6} \times 5$ $= 55 + \frac{5}{3}$ $= 55 + 1.666...$ $\approx 56.67$ kg. **The median weight of the students is 56.67 kg.**