RD Sharma Solutions Class 10 Statistics Exercise 7.4

RD Sharma Class 10 Solutions Chapter 7 Ex 7.4 PDF Free Download

Exercise 7.4

1. Following are the lives in hours of 15 pieces of the components of an aircraft engine. Find the median: 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Soln:

Given,

Lives of 15 component pieces (in hours) are given as:

= 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Now, to find median, the above data has to be arranged in ascending order.

= 694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

From the data, N = 15 (odd)

Median = $\left ( \frac{N + 1}{2} \right )^{th}$ terms

= $\left ( \frac{15 + 1}{2} \right )^{th}$ terms = 8th terms = 716

2. The following is the distribution of height of students of a certain class in a certain city:

 Height (in cm): 160-162                163-165                166-168                169-171                172-174 No of students: 15                           118                         142                         127                         18

Find the median height.

Soln:

The cumulative frequencies can be found out with their respective class intervals as given in the table below:

 Class interval (exclusive) Class interval  (inclusive) Class interval frequency Cumulative frequency 160 – 162 159.5 – 162.5 15 15 163 – 165 162.5 – 165.5 118 133 (F) 166 – 168 165.5 – 168.5 142 (f) 275 169 – 171 168.5 – 171.5 127 402 172 – 174 171.5 – 174.5 18 420 N = 420

It is given that N = 420

So, N/2 = 420/ 2 = 120

From the above table, the cumulative frequency which is just greater than N/2 is 275.

Now, 165.5 – 168.5 is the median class and,

f = 142,

L = 165.5,

F = 133 and

h = 168.5 – 165.5 = 3

Using the formula for median

Median= $L + \frac{\frac{N}{2}-F}{f}\times h$

= $165.5 + \frac {-39}{142}$

∴ Median = 166.5 + 1.63 = 168.13

3. Following is the distribution of I.Q of 100 students. Find the median I.Q.

 I.Q: 55-64     65-74     75-84     85-94     95-104     105-114     115-124    125-134      135-144 No of students: 1              2              9              22           33              22                 8                     2              1

Soln:

Make a table as follows:

 Class interval (exclusive) Class interval  (inclusive) Class interval frequency Cumulative frequency 55 – 64 54.5 – 64-5 1 1 65 – 74 64.5 – 74.5 2 3 75 – 84 74.5 – 84.5 9 12 85 – 94 84.5 – 94.5 22 34 (f) 95 – 104 94.5 – 104.5 33 (f) 67 105 – 114 104.5 – 114.5 22 89 115 – 124 114.5 – 124.5 8 97 125 – 134 124.5 – 134.5 2 99 135 – 144 134.5 – 144.5 1 100 N = 100

It is given that N = 100

So, N/ 2 = 100/ 2 = 50

From the above data, the cumulative frequency which is just greater than N/ 2 is 67.

Now, the median class is 94.5-104.5

F = 33

L=94.5, and

h =104.5 – 94.5=10

Now using the median formula:

Median = $L + \frac{\frac{N}{2}-F}{f}\times h$

= $94.5 + \frac{50-34}{33}\times 10$

∴ Median = 94.5 + 4.88 = 99.35

4.Calculate the median from the following data:

 Rent (in Rs): 15-25     25-35     35-45     45-55     55-65     65-75     75-85     85-95 No of houses: 8              10           15           25           40           20           15           7

Soln:

Again make a table similar to the previous questions:

 Class interval Frequency Cumulative frequency 15 – 25 8 8 25 – 35 10 18 35 – 45 15 33(f) 45 – 55 25 58 55 – 65 40(f) 28 65 – 75 20 38 75 – 85 15 183 85 – 95 7 140 N = 140

From the question, N = 140

So, N/ 2 = 140/ 2 = 70

From the data table, it is seen that the cumulative frequency which is just greater than N/ 2 is 98.

Now, the median class is 55 – 65 and

f = 40,

L = 55,

F = 58, and

h = 65 – 55 = 10

Now, using the median formula

Median = $L + \frac{\frac{N}{2}-F}{f}\times h$

= $55 + \frac{70 – 58}{40}\times 10$

∴ Median = 55 + 3 = 58

5.Calculate the median from the following data:

 Marks below: 10           20           30           40           50           60           70           80 No of students: 15           35           60           84           96           127         198         250

Soln:

 Marks below No of students Class interval Frequency Cumulative frequency 10 15 0 – 10 15 15 20 35 10 – 20 20 35 30 60 20 – 30 25 60 40 84 30 – 40 24 84 50 96 40 – 50 12 96(F) 60 127 50 – 60 31 (f) 127 70 198 60 – 70 71 198 80 250 70 – 80 52 250 N = 250

It is given that, N = 250

So, N/ 2 = 250/ 2 = 125

From the table, the cumulative frequency just greater than N/ 2 is 127

So, the median class is 50 – 60 and

f = 31,

L = 50,

F = 96,

h = 60 -50 = 10

Now, using the median formula

Median = $L + \frac{\frac{N}{2}-F}{f}\times h$

= $50 + \frac{125-96}{31}\times 10$

∴ Median= 50 + 9.35 = 59.35

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

 Age in years: 0-10       10-20     20-30     30-40     40-50 No of persons: 5              25           ?              18           7

Soln:

First, create the table as below:

 Class interval Frequency Cumulative frequency 0 – 10 5 5 10 – 20 25 30 (F) 20 – 30 x (f) 30 + x 30 – 40 18 48 + x 40 – 50 7 55 + x N = 170

It is given that,

Median = 24

From the table, median class = 20 – 30

So, F = 30, L = 20, h = 30 -20 = 10.

Let f = x

We know,

Median = $L + \frac{\frac{N}{2}-F}{f}\times h$

So, 24 = $20 + \frac{\frac{55+x}{2}-30}{x}\times 10$

24 – 20 = $\frac{\frac{55+x}{2}-30}{x}\times 10$

4x = $\left(\frac{55+x}{2}-30\right)\times 10$

4x = 275 + 5x – 300

4x – 5x = -25

-x = -25

x = 25

∴ The missing frequency is 25.

7. The following table gives the frequency distribution of married women by age at marriage.

 Age (in years) Frequency Age (in years) Frequency 15 – 19 53 40 – 44 9 20 – 24 140 45 – 49 5 25 – 29 98 50 – 54 3 30 – 34 32 55 – 59 3 35 – 39 12 60 and above 2

Calculate the median and interpret the results.

Soln:

Create tables similar to the previous questions:

 Class interval (exclusive) Class interval (inclusive) Frequency Cumulative frequency 15 – 19 14.5 – 19.5 53 53 (F) 20 – 24 19.5 – 24.5 140 (f) 193 25 – 29 24.5 – 29.5 98 291 30 – 34 29.5 – 34.5 32 323 35 – 39 34.5 – 39.5 12 335 40 – 44 39.5 – 44.5 9 344 45 – 49 44.5 – 49.5 5 349 50 – 54 49.5 – 54.5 3 352 55 – 54 54.5 – 59.5 3 355 60 and above 59.5 and above 2 357 N = 357

Given, N = 357

So, N/ 2 = 357/ 2 = 178.5

As the cumulative frequency just greater than N/ 2 is 193,

Then the median class is 19.5 – 24.5

l = 19.5,

F = 53,

f = 140,

h = 25.5 – 19.5 = 5

Median = $l + \frac{\frac{N}{2}-F}{f}\times h$

Median = $19.5 + \frac{178.5-53}{140}\times 5$

Median = 23.98

From the data, almost half the women were married between the age of 15 and 25.

8. The following table gives the distribution of the life time of 400 neon lamps:

 Life time: Number of lamps 1500 – 2000 14 2000 – 2500 56 2500 – 3000 60 3000 – 3500 86 3500 – 4000 74 4000 – 4500 62 4500 – 5000 48

Find the median life.

Soln:

The cumulative frequencies can be found out with their respective class intervals as given in the table below:

 Life time Number of lamps fi 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

From the data, it can be seen that the cumulative frequency which is just greater than n/ 2 (i.e. 400/ 2 = 200) is 216 which belongs to the class interval 3000 – 3500

So, Median class = 3000 – 3500

Frequency (f) of median class = 86

Lower limits (l) of median class = 3000

Cumulative frequency (cf) of class preceding median class = 130

Class size (h) = 500

Now, using the median formula,

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

= $3000 + \left (\frac{200 -130}{86}\right) \times 500 )$

= 3000 + (35000/ 86)

∴ Median = 3406.98

Hence, the median life time of these lamps is 3406.98 hours.

9.The distribution below gives the weight of 30 students in a class. Find the median weight of students:

 Weight (in kg): 40-45     45-50     50-55     55-60     60-65     65-70     70-75 No of students: 2              3              8              6              6              3              2

Soln:

Make the cumulative frequency table with their respective class intervals as given below:

 Weight (in kg) Number of students fi Cumulative frequency (cf) 40 – 45 2 2 45 – 50 3 5 50 – 55 8 13 55 – 60 6 19 60 – 65 6 25 65 – 70 3 28 70 – 75 2 30

From the table, cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19.

Now, this belongs to the class interval 55 – 60

So,

Median class = 55 – 60

l = 55

f = 6

Cumulative frequency (cf) = 13

h = 5

Now, using median formula,

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

= $55 + \left (\frac{15 -13}{6}\right) \times 5 )$

= 55 +10/ 6

= 56.666

∴ Median weight = 56.67 kg

10. Find the missing frequencies and the median for the following distribution if the mean is 1.46

 No. of accidents: 0              1              2              3              4              5 Total Frequencies (no of days): 46           ?              ?              25           10           5 200

Soln:

Make the table as given below:

 No of accidents (x) No of days (f) fx 0 46 0 1 a a 2 y 2b 3 25 75 4 10 40 5 5 25 N = 200 Sum = a + 2b + 140

From the data given, N = 200

So,

46 + a + b + 25 + 10 + 5 = 200

a + b = 200 – 46 – 25 – 10 – 5

⇒ a + b = 114 —- (Equation 1)

Also, Mean = 1.46

Sum/ N = 1.46

Now,

(a + 2b + 140)/ 200 = 1.46

a + 2b = 292 – 140

a + 2b = 152 —- (Equation 2)

Now, subtract Equation 1 from Equation 2

a + 2b – a – b = 152 – 114

b = 38

Now, put the value of b in Equation 1,

i.e. a = 114 – 38 = 76

After getting the values of variables a and b, make the cumulative frequency table.

 No of accidents No of days Cumulative frequency 0 46 46 1 76 122 2 38 160 3 25 185 4 10 195 5 5 200 N = 200

Here, N = 200

So, N/ 2 = 200/ 2 = 100

∴ The cumulative frequency which is just more than (N/ 2) is 122 then the median is 1.

11. An incomplete distribution is given below:

 Variable: 10-20     20-30     30-40     40-50     50-60     60-70     70-80 Frequency: 12           30           ?              65           ?              25           18

You are given that the median value is 46 and the total number of items is 230.

(i) Using the median formula fill up the missing frequencies.

(ii) Calculate the AM of the completed distribution.

Soln:

(i)

 Class interval Frequency Cumulative frequency 10 – 20 12 12 20 – 30 30 42 30 – 40 x 42+ x (F) 40 – 50 65 (f) 107 + x 50 – 60 Y 107 + x + y 60 – 70 25 132 + x + y 70 – 80 18 150 + x + y N = 150

In the question, Median = 46

Then, median class = 40 – 50

∴ f = 65, L = 40, h = 50 – 40 = 10, and F = 42 + x

Now, using the median formula

Median = $L+\frac{\frac{N}{2}-F}{f}\times h$

46 = $40+\frac{115-\left(42+x\right)}{65}\times 10$

46 – 40 = $\frac{115-42-x}{65}\times 10$

6 (65/ 10) = 73 –x

39 = 73 – x

∴ x = 73 – 39 = 34

It is also given that N = 230

⇒ 12 + 30 + 34 + 65 + y + 25 + 18 = 230

184 + y = 230

⇒ y = 230 – 184

⇒ y = 46

Thus the values of x and y are found out to be 34 and 46 respectively. The table will be as follows:

 Class interval Frequency Cumulative frequency 10 – 20 12 12 20 – 30 30 42 30 – 40 34 42+ 34 (F) 40 – 50 65 (f) 107 + 34 50 – 60 Y 107 + 34 + 46 = 187 60 – 70 25 132 + 34 +46 = 212 70 – 80 18 150 + 34 + 46 = 230 N = 150

(ii)

 Class interval Mid value x Frequency f Fx 10 – 20 15 12 180 20 – 30 25 30 750 30 – 40 35 34 1190 40 – 50 45 65 2925 50 – 60 55 46 2530 60 – 70 65 25 1625 70 – 80 75 18 1350 N = 230 Σ f x = 10550

Now, using the mean formula,

Mean = $\frac{\Sigma fx}{N}$

∴ Mean = 10550/ 230 = 45.87.

12. If the median of the following frequency distribution is 28.5 find the missing frequencies:

 Class interval: 0-10       10-20     20-30     30-40     40-50     50-60 Total Frequency: 5              f1             20           15           f2             5 60

Soln:

 Class interval Frequency Cumulative frequency 0 – 10 5 5 10 – 20 f1 5 + f1 (F) 20 – 30 20 (f) 25 + f1 30 – 40 15 40 + f1 40 – 50 f2 40 + f1 + f2 N = 60

It is given that Median = 28.5

Then, median class = 20 – 30

Using median formula,

Median = $l+\frac{\frac{N}{2}-F}{f}\times h$

⇒ 28.5 = $20\frac{30- \left( 5+ f_{1}\right)}{20}\times 10$

⇒ 28.5 -20 = $\frac{30- 5_{1}}{20}\times 10$

⇒8.5 = $\frac{25 –f_{1}}{2}$

⇒ 17 = 25 – f1

∴ f1 = 25 – 17 = 8

It is also given that,

Sum of frequencies = 60

So, 5 + f1 + 20 + 15 + f2 + 5 = 60

⇒ 5 + 8 + 20 + 15 + f2 + 5 = 60

Or, f2 = 7

Hence, the values f1 is 8 and f2 is 7.

13. The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data.

 Class interval Frequency Class interval Frequency 0 – 100 2 500 – 600 20 100 – 200 5 600 – 700 f2 200 – 300 f1 700 – 800 9 300 – 400 12 800 – 900 7 400 – 500 17 900 – 1000 4

Soln:

Creating a similar table,

 Class interval Frequency Cumulative frequency 0 – 100 2 2 100 – 200 5 7 200 – 300 f1 7 + f1 300 – 400 12 19 + f1 400 – 500 17 36 + f­1 (F) 500 – 600 20 (f) 56 + f1 600 – 700 f2 56 + f1 + f2 700 – 800 9 65 + f1 + f2 800 – 900 7 72 + f1 + f2 900 – 1000 4 76 + f1 + f2 N = 100

Given that,

Median = 525

Now, median class = 500 – 600

So, f = 20,

L = 500,

F = 36 + f1,  and

h = 600 – 500 = 100

We know, Median = $L+\frac{\frac{N}{2}-F}{f}\times h$

⇒ 525 = $500+\frac{50-\left( 36+ f_{1}\right)}{20}\times 100$

⇒ 525 = 500 +$\frac{50- 36- f_{1}}{20}\times 100$

⇒ 25 = (14 – f1) x 5

⇒ 5 = 14 – f1

∴ f1 = 14 – 5 = 9

It is also given that,

Sum of frequencies = 100

∴ 2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

⇒ 2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

∴ 85 + f2 = 100

Or, f2 = 100 – 85 = 15

Hence the values are f1 = 9 and f2 = 15

14. If the median of the following data is 32.5, find the missing frequencies.

 Class interval: 0-10       10-20     20-30     30-40     40-50     50-60     60-70 Total Frequency: f1             5              9              12           f2                   3              2 40

Soln:

 Class interval Frequency Cumulative frequency 0 – 10 f1 f1 10 – 20 5 5 + f1 20 – 30 9 14 + f1 30 – 40 12 (f) 26 + f1 40 – 50 f2 26 + f1 + f2 50 – 60 3 29 + f1 + f2 60 – 70 2 31 + f1 + f2 N = 40

It is given that Median = 32.5

So, median class = 90 – 40

Hence,

f = 12,

F = 14 + f1,

L = 30, and

h = 40 – 30 = 10,

Now, median = $L + \frac{\frac{N}{2}-F}{f}\times h$

⇒ 32.5 = $30 + \frac{20- \left( 14+ f_{1}\right)}{12}\times 10$

⇒ 32.5 – 30 = $\frac{20- \left( 14+ f_{1}\right)}{12}\times 10$

⇒ 2.5 (12) = (6 – f1) × 10

⇒ 30 = (6 – f1) × 10

Or, 3 = 6 – f1

∴ f1 = 6 – 3 = 3

Again, it is given that sum of frequencies = 40

So, f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40

⇒ 3 + 5 + 9 + 12 + f2 + 3 + 2 = 40

Or, 34 + f2 = 40

∴ f2 = 40 – 34 = 6

So, f1 = 3 and f2 = 6

15.Compute the median for each of the following data

 (i) (ii) Marks No of students Marks No of students Less than 10 0 More than 80 150 Less than 30 10 More than 90 141 Less than 50 25 More than 100 124 Less than 70 43 More than 110 105 Less than 90 65 More than 120 60 Less than 110 87 More than 130 27 Less than 130 96 More than 140 12 Less than 150 100 More than 150 0

Soln:

(i)

 Marks No of students Class interval Frequency Cumulative frequency Less than 10 0 0 – 10 0 0 Less than 30 10 10 – 30 10 10 Less than 50 25 30 ­– 50 15 25 Less than 70 43 50 – 70 18 43 (F) Less than 90 65 70 – 90 22 (f) 65 Less than 110 87 90 – 110 22 87 Less than 130 96 110 – 130 9 96 Less than 150 100 130 – 150 4 100 N = 100

Given,

N = 100

⇒ N/ 2 = 100/ 2 = 50

∴ median class = 70 – 90 (Since the cumulative frequency which is just greater than N/ 2 is 65)

So,

f = 22,

L = 70,

F = 43, and

h = 90 – 70 = 20

We know,

Median = $L + \frac{\frac{N}{2}-F}{f}\times h$

= $70 + \frac{50- 43}{22}\times 20$

= $70 + \frac{7\times 20}{22}$

= 70 + 6.36

= 76.36

(ii)

 Marks No of students Class interval Frequency Cumulative frequency More than 80 150 80 – 90 9 9 More than 90 141 90 – 100 17 26 More than 100 124 100 – 110 19 45 (F) More than 110 105 110 – 120 45 (f) 90 More than 120 60 120 – 130 33 123 More than 130 27 130 – 140 15 138 More than 140 12 140 – 150 12 150 More than 150 0 150 – 160 0 150 N  = 150

Here,

N = 150

⇒ N/ 2 = 150/ 2 = 75

∴ Median class is 110 – 120 (Since the cumulative frequency just more than N/ 2 is 90)

So,

f = 45,

L = 70,

F = 45, and

h = 120 – 110 = 10

Now,

Median = $L + \frac{\frac{N}{2}-F}{f}\times h$

= $110 + \frac{75- 45}{45}\times 10$

= $110 + \frac{30\times 10}{45}$

= 110 + 6.67

= 116.67

16. A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data were obtained:

 Height in cm number of girls Less than 140 Less than 145 Less than 150 Less than 155 Less than 160 Less than 165 4 11 29 40 46 51

Find the median height.

Soln:

First, it is important to find the class intervals and their corresponding frequencies for calculating the median height.

Now, the given distribution is of the less than type. So, 140, 145, 150, 155, 160, and 165 gives the upper limits of the corresponding class intervals.

Hence, the classes should be below 140, 140-145, 145-150, 150-155, 155-160, 160-165.

From the given data, it can be observed that there are 4 girls with height less than 140 which means the frequency of class interval below 140 is 4.

So, it can be concluded that there are 11 girls with heights less than 145 and 4 girls with height less than 140.

Hence, the total number of girls whose heights are in the interval of (140 – 145) is 11 – 4 = 7.

Also, the frequency of 145 – 150 is 29 – 11 = 18,

For 150 – 155, it is 40 – 29 = 11, and so on.

∴ The frequency distribution table becomes:

 Class interval Frequency Cumulative frequency Below 140 140 – 145 145 – 150 150 – 155 155 – 160 160 – 165 4 7 18 11 6 5 4 11 29 40 46 51

It is given that n = 51.

S0, n/ 2 = 51/ 2 =25.5

Now, it can be seen that this observation lies in the class 145 – 150

So,

f (the frequency of the median class 145 – 150) = 18

L (the lower limit) = 145

h (the class size) = 5

cf (the cumulative frequency of the class preceding 145 – 150) = 11

Now, median = $l + \left ( \frac{\frac{n}{2}-cf}{f} \right ) \times h$, we have

= 145 + 72.5/ 18 = 149.03

So, 149.03 is the median height of the girls.

From this data, it can be concluded that 50% of the girls have less than this height while and 50% are taller than this height.

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

 Ages in years Number of policy holders Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60 2 6 24 45 78 89 92 98 100

Soln:

By looking at the data, it can be seen that the class width is not the same in this case. So, we need not adjust the frequencies according to the class interval.

It is also noticed that the given frequency table is of less than type and is represented with upper-class limits.

Now, as policies are given only to the persons whose ages are between 18 and 60 years, the respective cumulative frequency is as below.

 Age (in years) Number of policy holders fi Cumulative frequency (cf) 18 – 20 2 2 20 – 25 6 – 2 = 4 6 25 – 30 24 – 6 = 18 24 30 – 35 45 – 24 = 21 45 35 – 40 78 – 45 = 33 78 40 – 45 89 – 78 = 11 89 45 – 50 92 – 89 = 3 92 50 – 55 98 – 92 = 6 98 55 – 60 100 – 98 = 2 100 Total

Now, n = 100

Or, n/2 = 50

So, “cf” which is just greater than 50 is 78 which belongs to the interval 35 – 40

∴ Median class = 35 – 40

Now,

Class size (h) = 5

Lower limit (l) of median class = 35

Cumulative frequency (cf) of class preceding median class = 45 and

Frequency (f) of median class = 33

We know,

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

= $35 + \left ( \frac{50 – 45}{33}\right )\times 5$

= $35 + \frac {25}{33}$

= 35.76

∴ The median age is 35.76 years.

18. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

 Length (in mm) No of leaves 118 – 126 3 127 – 135 5 136 – 144 9 145 – 153 12 154 – 162 5 163 – 171 4 172 – 180 2

Find the mean length of life

Soln:

From the data, it can be noticed that it is not having continuous class intervals. Here, the difference between the two class intervals is 1.

So, we have to add and subtract 1/ 2 = 0.5 to upper class limits and lower class limits.

Now, the continuous class intervals with respective cumulative frequencies can be represented as below:

 Length (in mm) Number of leaves fi Cumulative frequency (cf) 117.5 – 126.5 3 3 126.5 – 135.5 5 8 135.5 – 144.5 9 17 144.5 – 153.5 12 29 153.5 – 162.5 5 34 162.5 – 171.5 4 38 171.5 – 180.5 2 40

From the table,

n/2 = 20

Now, the cumulative frequency which is just greater 20 is 29. So, it belongs to the class interval 144.5 – 153.5

∴ Median class = 144.5 – 153.5

Class size (h) = 9

Lower limit (l) = 144.5

Cumulative frequency (cf) of class preceding median class = 17

Frequency (f) of median class = 12

Now,

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

= $144.5 + \left (\frac{20 -17}{12}\right) \times 9 )$

= 144.5 + 9/ 4 = 146.75

∴ The Median length of leaves is 146.75 mm

19.An incomplete distribution is given as follows:

 Variable: 0-10       10-20     20-30     30-40     40-50     50-60     60-70 Frequency: 10           20           ?              40           ?              25           15

You are given that the median value is 35 and sum is all the frequencies are 170. Using the median formula, fill up the missing frequencies

Soln:

 Class interval Frequency Cumulative frequency 0 – 10 10 10 10 – 20 20 30 20 – 30 f1 30 + f1 (F) 30 – 40 40 (F) 70 + f1 40 – 50 f2 70 + f1 + f2 50 – 60 25 95 + f1 + f2 60 – 70 15 110 + f1 + f2 N = 170

Given

Median = 35

Then median class = 30 – 40

L = 30, h = 40 – 30 = 10, f = 40, F = 30 + f1

Median = $L+\frac{\frac{N}{2}-F}{f}\times h$

35 = $30+\frac{85-( 30+ f_{1})}{40}\times 10$

35 – 30 = $\frac{85–30–f_{1}}{40}\times 10$

5 = $\frac{55 – f_{1}}{4}$

20 = 55 – f1

f1 = 55 – 20 = 35

Given

Sum of frequencies = 170

10 + 20 + fi + 40 + f2 + 25 + 15 = 170

10 + 20 + 35 + 40 + f2 + 25 + 15 = 170

f2 = 25

f1 = 35 and f2 =25