Exercise 13.3 Practice
Median of Grouped Data & Missing Frequencies
Q1: Water Consumption
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The following frequency distribution gives the monthly consumption of water (in units) of 80 consumers of a locality. Find the median, mean and mode of the data and compare them.
| Consumption (units) | 50-70 | 70-90 | 90-110 | 110-130 | 130-150 | 150-170 | 170-190 |
|---|---|---|---|---|---|---|---|
| Consumers | 4 | 8 | 12 | 20 | 16 | 12 | 8 |
Median: $N=80, N/2=40$. CF of classes: 4, 12, 24, 44, 60, 72, 80.
Median class is 110-130. $l=110, cf=24, f=20, h=20$.
Median $= 110 + \frac{40-24}{20} \times 20 = 110 + 16 = 126$.
Median class is 110-130. $l=110, cf=24, f=20, h=20$.
Median $= 110 + \frac{40-24}{20} \times 20 = 110 + 16 = 126$.
Mode: Max freq = 20. Modal class 110-130.
Mode $= 110 + \frac{20-12}{40-12-16} \times 20 = 110 + \frac{8}{12} \times 20 = 110 + 13.33 = 123.33$.
Mode $= 110 + \frac{20-12}{40-12-16} \times 20 = 110 + \frac{8}{12} \times 20 = 110 + 13.33 = 123.33$.
Mean: $\Sigma f_i = 80$. Calculated mean $\approx 125$.
Median: 126; Mode: 123.33; Mean: 125 (approx).
Q2: Missing Values (x, y)
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If the median of the distribution given below is 30, find the values of x and y. Total frequency is 100.
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Frequency | 10 | x | 25 | 30 | y | 10 |
Given $N=100$, Median=30. Median lies in 30-40 (since it is 30, we use this class, or sometimes 20-30 if continuous. Given the structure, 30-40 is standard).
Sum of freq: $75 + x + y = 100 \Rightarrow x + y = 25$.
Median formula: $30 = 30 + \frac{50 - (35+x)}{30} \times 10$.
$0 = \frac{15-x}{3} \Rightarrow x = 15$.
$0 = \frac{15-x}{3} \Rightarrow x = 15$.
$15 + y = 25 \Rightarrow y = 10$.
x = 15, y = 10.
Q3: Age Policy
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An insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age.
| Age (in years) | Below 20 | Below 25 | Below 30 | Below 35 | Below 40 | Below 45 |
|---|---|---|---|---|---|---|
| No. of policy holders | 6 | 15 | 29 | 55 | 85 | 100 |
Convert to class intervals: 15-20 (6), 20-25 (9), 25-30 (14), 30-35 (26), 35-40 (30), 40-45 (15).
$N=100, N/2=50$. Median class is 30-35 (CF goes from 29 to 55).
$l=30, cf=29, f=26, h=5$.
Median $= 30 + \frac{50-29}{26} \times 5 = 30 + \frac{21}{26} \times 5 = 30 + 4.04$.
Median $= 30 + \frac{50-29}{26} \times 5 = 30 + \frac{21}{26} \times 5 = 30 + 4.04$.
Median Age is 34.04 years.
Q4: Length of Leaves
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The lengths of 50 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
Find the median length of the leaves. (Hint: Convert to continuous classes).
| Length (mm) | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 |
|---|---|---|---|---|---|
| No. of leaves | 5 | 10 | 15 | 12 | 8 |
Continuous Classes: 109.5-119.5, 119.5-129.5, 129.5-139.5, etc.
$N=50, N/2=25$. CF: 5, 15, 30, 42, 50. Median class is 129.5-139.5.
$l=129.5, cf=15, f=15, h=10$.
Median $= 129.5 + \frac{25-15}{15} \times 10 = 129.5 + 6.67$.
Median $= 129.5 + \frac{25-15}{15} \times 10 = 129.5 + 6.67$.
Median length is 136.17 mm.
Q5: Bulb Lifetimes
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The following table gives the distribution of the life time of 500 neon lamps:
Find the median life time of a lamp.
| Life time (hours) | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 |
|---|---|---|---|---|---|---|---|
| Number of lamps | 20 | 60 | 70 | 100 | 90 | 80 | 80 |
$N=500, N/2=250$. CFs: 20, 80, 150, 250, 340...
Wait, exactly 250 is the CF of class 3000-3500. So median class is 3000-3500? Or next?
Rule: If cumulative freq is exactly N/2, median is the upper limit. Let's calculate.
Actually, if CF reaches 250 at end of 3000-3500, then 250th item is the last one in that class. So Median $\approx 3500$.
Let's use formula: Class containing 250th item. 3000-3500 contains items 151 to 250.
$l=3000, cf=150, f=100, h=500$.
Median $= 3000 + \frac{250-150}{100} \times 500 = 3000 + \frac{100}{100} \times 500 = 3500$.
Median $= 3000 + \frac{250-150}{100} \times 500 = 3000 + \frac{100}{100} \times 500 = 3500$.
Median lifetime is 3500 hours.
Q6: Surnames
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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:
Determine the median number of letters. Find the mean number of letters. Also, find the modal size of the surnames.
| Number of letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 |
|---|---|---|---|---|---|---|
| Number of surnames | 8 | 35 | 45 | 8 | 2 | 2 |
Median: $N=100, N/2=50$. CF: 8, 43, 88... Median class 7-10.
Median $= 7 + \frac{50-43}{45} \times 3 = 7 + 0.47 = 7.47$.
Median $= 7 + \frac{50-43}{45} \times 3 = 7 + 0.47 = 7.47$.
Mode: Max freq = 45 (Class 7-10).
Mode $= 7 + \frac{45-35}{90-35-8} \times 3 = 7 + \frac{10}{47} \times 3 = 7.64$.
Mode $= 7 + \frac{45-35}{90-35-8} \times 3 = 7 + \frac{10}{47} \times 3 = 7.64$.
Mean: Approx 7.8.
Median: 7.47; Mode: 7.64; Mean: 7.8.
Q7: Student Weights
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The distribution below gives the weights of 40 students of a class. Find the median weight.
| Weight (kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
|---|---|---|---|---|---|---|---|
| Students | 3 | 4 | 10 | 8 | 8 | 4 | 3 |
$N=40, N/2=20$. CF: 3, 7, 17, 25... Median class 55-60.
$l=55, cf=17, f=8, h=5$.
Median $= 55 + \frac{20-17}{8} \times 5 = 55 + \frac{15}{8} = 55 + 1.875$.
Median weight is 56.875 kg.