Q1: In a hospital, the ages of diabetic patients were recorded as follows

Age (in years) | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 |

Number of patients | 5 | 20 | 40 | 50 | 25 |

Find the median age.

Sol:

We prepare the cumulative frequency table, as shown:

Age (in years) | Number of patients | Cumulative frequency |

0-15 | 5 | 5 |

15-30 | 20 | 25 |

30-45 | 40 | 65 |

45-60 | 50 | 115 |

60-75 | 25 | 140 |

Total | \(N= \Sigma f_{i}= 140\) |

Now, N= 140

\(\Rightarrow \frac{N}{2} = 70\)

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45-60.

Thus the median class is 45-60.

We know,

\(Median = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

And l= 45, h= 15, f= 50, N= 140 and cf= 65

\(Median = 45+\left ( \frac{\frac{140}{2}-65}{50} \right )\times 15\)

\(Median = 45+\left ( \frac{70-65}{50} \right )\times 15\)

\(Median = 45 + 1.5\)

\(Median = 46.5\)

Hence the median age is 46.5 years.

Q2: Compute the median from the following data :

Marks | 0-7 | 7-14 | 14-21 | 21-28 | 28-35 | 35-42 | 42-49 |

Number of students | 3 | 4 | 7 | 11 | 0 | 16 | 9 |

Sol:

We prepare the cumulative frequency table, as shown:

Marks | Number of students
(f) |
Cumulative frequency |

0-7 | 3 | 3 |

7-14 | 4 | 7 |

14-21 | 7 | 14 |

21-28 | 11 | 25 |

28-35 | 0 | 25 |

35-42 | 16 | 41 |

42-49 | 9 | 50 |

\(N= \Sigma f = 50\) |

N= 50

\(\Rightarrow \frac{N}{2} = 25\)

The cumulative frequency greater than 25 is 41 and the corresponding class is 35-42.

Thus the median class is 35-42.

Therefore, l=35, h = 7, f = 16, Â c.f. of preceding class = 25 and \( \frac{N}{2} = 25\)

\(Median = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\(Median = 35 + \left ( \frac{25-25}{16} \right )\times 7\)

\(Median = 35+ 0\)

\(Median = 35 \)

Q3: The following tables shows the daily wages of workers in a factory:

Daily wages (in Rs.) | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |

Number of workers | 40 | 32 | 48 | 22 | 8 |

Find the median daily wage income of the workers.

Sol

We prepare the cumulative frequency table, as shown:

Daily wages | Number of workers | Cumulative frequency |

0-100 | 40 | 40 |

100-200 | 32 | 72 |

200-300 | 48 | 120 |

300-400 | 22 | 142 |

400-500 | 8 | 150 |

\(N= \Sigma f = 150\) |

N= 150

\(\Rightarrow \frac{N}{2} = 75\)

The cumulative frequency greater than 75 is 120 and the corresponding class is 200-300.

Thus the median class is 200-300.

Therefore, l=200, h = 100, f = 48, c.f. of preceding class = 75 and \( \frac{N}{2} = 75\)

\(Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\(Median = 200 + \left ( \frac{75-72}{48} \right )\times 100 \)

\(Median = 200+ 6.25\)

\(Median = 206.25 \)

Hence, the median daily wage income of the worker is Rs. 206.25

Q4: Calculate the median from the following frequency distribution:

Class | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 |

Frequency | 5 | 6 | 15 | 10 | 5 | 4 | 2 | 2 |

Sol:

We prepare the cumulative frequency table, as shown:

Class | Frequency
(f) |
Cumulative frequency |

5-10 | 5 | 5 |

10-15 | 6 | 11 |

15-20 | 15 | 26 |

20-25 | 10 | 36 |

25-30 | 5 | 41 |

30-35 | 4 | 45 |

35-40 | 2 | 47 |

40-45 | 2 | 49 |

\(N= \Sigma f = 49\) |

N= 49

\(\Rightarrow \frac{N}{2} = 24.5 \)

The cumulative frequency greater than 24.5 is 26 and the corresponding class is 15-20.

Thus the median class is 15-20.

Therefore, l= 15, h = 5, f = 15, Â c.f. of preceding class = 11 and \( \frac{N}{2} = 24.5 \)

\(Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\(Median = 15 + \left ( \frac{24.5 – 11}{15} \right )\times 5 \)

\(Median = 15 + 4.5\)

\(Median = 19.5 \)

Hence, median is 19.5.

Q5: Given below is the number of units of electricity consumed in a week in a certain locality :

Consumption (in units) | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 |

Number of consumption | 4 | 5 | 13 | 20 | 14 | 7 | 4 |

Sol:

We prepare the cumulative frequency table, as shown:

Class | Frequency
(f) |
Cumulative frequency |

65-85 | 4 | 4 |

85-105 | 5 | 9 |

105-125 | 13 | 22 |

125-145 | 20 | 42 |

145-165 | 14 | 56 |

165-185 | 7 | 63 |

185-205 | 4 | 67 |

\(N= \Sigma f = 67\) |

N= 67

\(\Rightarrow \frac{N}{2} = 33.5 \)

The cumulative frequency greater than 33.5 is 42 and the corresponding class is 125-145.

Thus the median class is 125-145.

Therefore, l= 125, h = 20, f = 20, Â c.f. of preceding class = 22 and \( \frac{N}{2} = 33.5 \)

\(Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\(Median = 125 + \left ( \frac{33.5 – 22}{20} \right )\times 20 \)

\(Median = 125 + 11.5\)

\(Median = 136.5 \)

Hence, median is 136.5 .

Q6: Calculate the median from the following data

Height | 135-140 | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 |

No. of boys | 6 | 10 | 18 | 22 | 20 | 15 | 6 | 3 |

Sol:

We prepare the cumulative frequency table, as shown/p>

Class | Frequency
(f) |
Cumulative frequency |

135-140 | 6 | 6 |

140-145 | 10 | 16 |

145-150 | 18 | 34 |

150-155 | 22 | 56 |

155-160 | 20 | 76 |

160-165 | 15 | 91 |

165-170 | 6 | 97 |

170-175 | 3 | 100 |

\(N= \Sigma f = 100\) |

N= 100

\(\Rightarrow \frac{N}{2} = 50 \)

The cumulative frequency greater than 50 is 56 and the corresponding class is 150-155.

Thus the median class is 150-155

Therefore, l= 150, h = 5, f = 22, Â c.f. of preceding class = 34 and \( \frac{N}{2} = 50 \)

\(Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\(Median = 150 + \left ( \frac{50 – 34}{22} \right )\times 5 \)

\(Median = 150 + 3.64\)

\(Median = 153.64 \)

Hence, median is 153.64.

Q7: Calculate the missing frequency from the distribution, it being given that the median of the distribution is 24.

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Frequency | 5 | 25 | ? | 18 | 7 |

Sol:

We prepare the cumulative frequency table, as shown:

Class | Frequency
(f) |
Cumulative frequency |

0-10 | 5 | 5 |

10-20 | 25 | 30 |

20-30 | x | x+30 |

30-40 | 18 | x+48 |

40-50 | 7 | x+55 |

Median is 24 which lies in 20-30

Therefore, Median Class = 20-30

Let the unknown frequency be x.

l= 20, h = 10, f = x, Â c.f. of preceding class = 30 and \( \frac{N}{2} = \frac{x+55}{2} \)

\(Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\(24 = 20 +\frac{\frac{x+55}{2}-30}{x}\times 10\)

\(\Rightarrow 24 = 20 +\frac{\frac{x+55-60}{2}}{x}\times 10\)

\(\Rightarrow 24 = 20 +\frac{x-5}{2x}\times 10\)

\(\Rightarrow 24 – 20= \frac{x-5}{x}\times 5\)

\(\Rightarrow 4x = 5x – 25\)

\(\Rightarrow x = 25\)

Hence, the unknown frequency is 25.

Q8: The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |

Frequency | 120 | a | 12 | 15 | b | 6 | 6 | 4 |

Sol:

We prepare the cumulative frequency table, as shown:

Class | Frequency\(f_{i}\) |
Cumulative frequency |

0-5 | 12 | 12 |

5-10 | a | 12+a |

10-15 | 12 | 24+a |

15-20 | 15 | 39+a |

20-25 | b | 39+a+b |

25-30 | 6 | 45+a+b |

30-35 | 6 | 51+a+b |

35-40 | 4 | 55+a+b |

Total | \(N= \Sigma f_{i} = 70\) |

Let a,b be the missing frequencies of class intervals 5-10 and 20-25 respectively. Then,

55 + a + b = 70

\(\Rightarrow a+b = 15\)

Median is 16, which lies in15-20. So the median class is 15-20.

Therefore, l= 15, h= 5, N= 70, f= 15, and cf = 24+a

Now,

\(Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\( 16 = 15 +\frac{\frac{70}{2}-(24 + a)}{15}\times 5 \)

\(\Rightarrow 16 = 15 + \frac{35 – 24 – a}{3} \)

\(\Rightarrow 16 – 15 = \frac{11-a} {3} \)

\(\Rightarrow 1 \times 3 = 11 – a \)

\(\Rightarrow a = 11 – 3 \)

\(\Rightarrow a = 8 \)

Therefore, \(b = 15 – a\)

\(\Rightarrow b = 15 – 8\)

\(\Rightarrow b = 7\)

Hence, a= 8 and b = 7.

Q9: In the following data the median of the runs scored by top 60 batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.

Runs scored | 2500-3500 | 3500-4500 | 4500-5500 | 5500-6500 | 6500-7500 | 7500-8500 |

No. of batsmen | 5 | x | y | 12 | 6 | 2 |

Sol:

We prepare the cumulative frequency table, as shown:

Runs scored | No. of batsmen | Cumulative frequency |

2500-3500 | 5 | 5 |

3500-4500 | x | 5+x |

4500-5500 | y | 5+x+y |

5500-6500 | 12 | 17+x+y |

6500-7500 | 6 | 23+x+y |

7500-8500 | 2 | 25+x+y |

Total | \(N= \Sigma f_{i} = 60\) |

Let x,y be the missing frequencies of class intervals 3500-4500 and 4500-5500 respectively.

Then,

25 + x + y = 60

\(\Rightarrow x+y = 35 \)

Median is 5000, which lies in 4500-5500.

So the median class is 4500-5500.

Therefore, l= 4500, h= 1000, N= 60, f= y, and cf = 5+x

Now,

\(Median, M = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\( 5000 = 4500 +\frac{\frac{60}{2} – (5 + x)}{y}\times 1000 \)

\(\Rightarrow 5000 – 4500 = \frac{30 – 5 – x}{y} \times 1000 \)

\(\Rightarrow 500 = \frac{25-x} {y} \times 1000 \)

\(\Rightarrow y = 50 – 2x \)

\(\Rightarrow 35 – x = 50 – 2x \)

\(\Rightarrow 2x – x = 50 – 35 \)

\(\Rightarrow x = 15 \)

Therefore, \(y = 35 – x\)

\(\Rightarrow y = 35 – 15 \)

\(\Rightarrow y = 20 \)

Hence, x= 15 and y = 20.

Q10: If the median of the following frequency distribution is 32.5, find the values of \(f_{1} \; and\; f_{2}\)

Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |

Frequency | \(f_{1}\) |
5 | 9 | 12 | \(f_{2}\) |
3 | 2 | 40 |

Sol:

We prepare the cumulative frequency table, as shown below:

Class | Frequency | Cumulative frequency |

0-10 | \(f_{1}\) |
\(f_{1}\) |

10-20 | 5 | \(f_{1}\) |

20-30 | 9 | \(f_{1}\) |

30-40 | 12 | \(f_{1}\) |

40-50 | \(f_{2}\) |
\(f_{1}\) |

50-60 | 3 | \(f_{1}\) |

60-70 | 2 | \(f_{1}\) |

\(N= \Sigma f = 40\) |

Now, \(f_{1}\)

\(\Rightarrow f_{1}+ f_{2} = 9\)

\(\Rightarrow f_{2} = 9 – f_{1}\)

The median is 32.5 which lies in 30 – 40.

Hence, median class = 30 – 40

Therefore, l= 30,\(\frac{N}{2} = 20\)

Now,

\(Median = 32.5 \; Also, \; Median (M) = l+\left ( \frac{\frac{N}{2}-cf}{f} \right )\times h\)

\( 32.5 = 30 + \frac{20 – (14 + f_{1})}{12} \times 10 \)

\(\Rightarrow 32.5 = 30 + \frac{6 – f_{1}}{12} \times 10 \)

\(\Rightarrow 2.5 = \frac{ 6 – f_{1} } {12} \times 10 \)

\(\Rightarrow 60 – 10f_{1} Â = 30 \)

\(\Rightarrow 10f_{1} = 30 \)

\(\Rightarrow f_{1} Â = 3 \)

From eq. (i), we have

\(f_{2}= 9-3\)

\(\Rightarrow f_{2}= 6\)

Q11:Calculate the median for the following data:

Age( in years) | 19-25 | 26-32 | 33-39 | 40-46 | 47-53 | 54-60 |

Frequency | 35 | 96 | 68 | 102 | 35 | 4 |

Answer:

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

Class | Frequency \(f_{i}\) |
C.F |

18.5-25.5 | 35 | 35 |

25.5-32.5 | 96 | 131 |

32.5-39.5 | 68 | 199 |

39.5-46.5 | 102 | 301 |

46.5-53.5 | 35 | 336 |

53.5-60.5 | 4 | 340 |

\(N=\sum f_{i}=340\) |

N =340

\(\Rightarrow \frac{N}{2} =170\)

Now,

The cumulative frequency just greater than 170 is 199 and corresponding class is 32.5-39.5.

Thus, the median class is 32.5-39.5.

I = 32.5, h = 7, f = 68, c.f. = cf preceding class is 131 and N/2 = 170

Median = \(I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 32.5+\left [ 7\times \frac{\left (170-131 \right )}{22} \right ]\)

= 32.5 + 4 = 36.5

Hence median is 36.5 years.

Q12: Find the median wages for the following frequency distribution:

Wages per day ( in Rs.) | 61-70 | 71-80 | 81-90 | 91-100 | 101-110 | 111-120 |

No. of workers | 5 | 15 | 20 | 30 | 20 | 8 |

Sol:

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get/p>

Wages per day(in Rs.) | Frequency \(f_{i}\) |
C.F |

60.5-70.5 | 5 | 5 |

70.5-80.5 | 15 | 20 |

80.5-90.5 | 20 | 40 |

90.5-100.5 | 30 | 70 |

100.5-110.5 | 20 | 90 |

110.5-120.5 | 8 | 98 |

\(N=\sum f_{i}=98\) |

N =98 => N/2 = 49

Now,

The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5-100.5.

Thus, the median class is 90.5-100.5

I = 90.5, h = 10, f = 30, c = C.F. preceding median class = 40 and N/2 = 49

Median = \(I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 90.5+\left [ 10\times \frac{\left (49-40 \right )}{30} \right ]\)

= 90.5+3 = Rs. 93.5

Hence, median is Rs. 93.5

Q13: Find the median from the following data:

Class | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 |

Frequency | 7 | 10 | 16 | 32 | 24 | 16 | 11 | 5 | 2 |

Sol:

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

Marks | Frequency \(f_{i}\) |
C.F |

0.5-5.5 | 7 | 7 |

5.5-10.5 | 10 | 17 |

10.5-15.5 | 16 | 33 |

15.5-20.5 | 32 | 65 |

20.5-25.5 | 24 | 89 |

25.5-30.5 | 16 | 105 |

30.5-35.5 | 11 | 116 |

35.5-40.5 | 5 | 121 |

40.5-45.5 | 2 | 123 |

\(N=\sum f_{i}=123\) |

N =123

=> N/2 = 61.5

Now,

The cumulative frequency just greater than 61.5 is 65 and corresponding class is 15.5-20.5.

Thus, the median class is 15.5-20.5

I = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33 and N/2 = 61.5

Median = \(I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 15.5+\left [ 5\times \frac{\left (61.5-33 \right )}{32} \right ]\)

= 15.5+4.45 = 19.95

Hence, median = 19.95

Q14: Find the median from the following data:

Marks | Frequency \(f_{i}\) |

Below 10 | 12 |

Below 20 | 32 |

Below 30 | 57 |

Below 40 | 80 |

Below 50 | 92 |

Below 60 | 116 |

Below 70 | 164 |

Below 80 | 200 |

Sol:

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

Marks | Frequency \(f_{i}\) |
C.F |

0-10 | 12 | 12 |

10-20 | 20 | 32 |

20-30 | 25 | 57 |

30-40 | 23 | 80 |

40-50 | 12 | 92 |

50-60 | 24 | 116 |

60-70 | 24 | 116 |

70-80 | 36 | 200 |

\(N=\sum f_{i}=200\) |

N =200

=> N/2 = 100

Now,

The cumulative frequency just greater than 100 is 116 and corresponding class is 50-60.

Thus, the median class is 50-60

I = 50, h = 10, f = 24, c = C.F. preceding median class = 92 and N/2 = 100

Median = \(I+\left [ h\times \frac{\left ( \frac{N}{2}-c \right )}{f} \right ]= 50+\left [ 10\times \frac{\left (100-92 \right )}{24} \right ]\)

\(=50+\left [ 10\times \frac{84}{24} \right ]\)

= 50 + 3.33 = 53.33

Hence, Median = 53.33