Ex-7.4 Statistics, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

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

Sol:Lives in hours of is pieces are = 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Arrange the above data in ascending order = 694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

N = 15 (odd)

Median =  terms

 terms = 8^th terms = 716

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

Find the median height.

Sol:

We have

N = 420

N/2 = 420/ 2 = 120

The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class such, that

L = 165.5, f = 142, F = 133 and h = 168.5 – 165.5 = 3

Mean = 

= Missing close brace = 166.5 + 1.63 = 168.13

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

Sol:

We have N = 100

N/ 2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is 94.5-104.5 such that

L=94.5,F = 33 h =104.5 – 94.5=10

Mean = 

= 94.5+×10 = 94.5 + 4.88 = 99.35

4.Calculate the median from the following data:

Sol:

We have N = 140

N/ 2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 such that

L = 55, f = 40, F = 58, h = 65 – 55 = 10

Mean = 

= 55+ = 55 + 3 = 58

5.Calculate the median from the following data:

Sol:

We have N = 250

N/ 2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 such that

L = 50, f = 31, F = 96, h = 60 -50 = 10

Mean = 

= 50+ = 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.

Sol:

Given

Median = 24

Then, median class = 20 – 30

L = 20,   h = 30 -20 = 10, f = x,      F = 30

Median = l + (N/2 − F)/f × h

4x = 275 + 5x – 300

4x – 5x = -25

-x = -25

x = 25

Missing frequency = 25

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

Calculate the median and interpret the results

Sol:

N = 357

N/ 2 = 357/ 2 = 178.5

The cumulative frequency just greater than N/ 2 is 193,

Then the median class is 19.5 – 24.5 such that l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

Median = l + (N/2 − F)/f × h

Median = 19.5 + 

Median = 23.98

Nearly half the women were married between the ages of 15 and 25

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

Find the median life.

Sol: We can find cumulative frequencies with their respective class intervals as below

Now we may observe that cumulative frequency just greater than n/ 2 (400/ 2 = 200) is 216 belongs to class interval 3000 – 3500

Median class = 3000 – 3500

Lower limits (l) of median class = 3000

Frequency (f) of median class = 86

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

Class size (h) = 500

Median = 

= 3000+

= 3000 + (35000/ 86)

= 3406.98

So, median life time of lamps is 3406.98 hours

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

Sol: We may find cumulative frequency with their respective class intervals as below

Cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belonging to class interval 55 – 60

Median class = 55 – 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) = 13

Class size (h) = 5

Median = 

= 55+ 

= 55 +10/ 6

= 56.666

So, median weight is 56.67 kg

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

Sol:

Given

N = 200

46 + x + y + 25 + 10 + 5 = 200

x + y = 200 – 46 – 25 – 10 – 5

x + y = 114 —- (1)

And, Mean = 1.46

Sum/ N = 1.46

(x + 2y + 140)/ 200 = 1.46

x + 2y = 292 – 140

x + 2y = 152 —- (2)

Subtract equation (1) from equation (2)

x + 2y – x – y = 152 – 114

y = 38

Putting the value of y in equation (1), we have x = 114 – 38 = 76

We have,

N = 200

N/ 2 = 200/ 2 = 100

The cumulative frequency just more than N/ 2 is 122 then the median is 1

11.An incomplete distribution is given below:

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.

Sol:

(i)

Given

Median = 46

Then, median class = 40 – 50

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

Median = 

6 (65/ 10) = 73 –x

39 = 73 – x

x = 73 – 39 = 34

Given

N = 230

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

184 + y = 230

Y = 230 – 184

Y = 46

(ii)

Mean = 

= 10550/ 230 = 45.87

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

Sol:

Given

Median = 28.5

Then, median class = 20 – 30

Median = 

17 = 25 – f_{1}

f₁ = 25 – 17 = 8

Given

Sum of frequencies = 60

5 + f₁ + 20 + 15 + f₂ + 5 = 60

5 + 8 + 20 + 15 + f₂ + 5 = 60

f₂ = 7

f₁ = 8 and f₂ = 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.

Sol:

Given

Median                = 525

Then, median class = 500 – 600

L = 500, f = 20, F = 36 + f₁, h = 600 – 500 = 100

Median =

25 = (14 – f₁) x 5

5 = 14 – f₁

f₁ = 14 – 5 = 9

Given

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

f₂ = 100 – 85 = 15

f₁ = 9 and f₂ = 15

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

Sol:

Given

Median = 32.5

The median class = 90 – 40

L = 30, h = 40 – 30 = 10, f = 12, F = 14 + f₁

Median = 

2×10

2.5 (12) = (6 – f₁) * 10

30 = (6 – f₁) * 10

3 = 6 – f₁

f₁ = 6 – 3 = 3

Given

Sum of frequencies = 40

f₁ + 5 + 9 + 12 + f₂ + 3 + 2 = 40

3 + 5 + 9 + 12 + f₂ + 3 + 2 = 40

34 + f₂ = 40

f₂ = 40 – 34 = 6

f₁ = 3 and f₂ = 6

15.Compute the median for each of the following data

Sol:(i)

We have

N = 100

N/ 2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 65 then median class is 70 – 90 such that

L = 70, h = 90 – 70 = 20, f = 22, F = 43

Median = 

= 70 + 6.36

= 76.36

(ii)

We have

N = 150

N/ 2 = 150/ 2 = 75

The cumulative frequency just more than N/ 2 is 90 then the median class is 110 – 120 such that

L = 70, h = 120 – 110 = 10, f = 45, F = 45

Median =

= 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 was obtained:

Find the median height.

Sol: To calculate the median height, we need to find the class intervals and their corresponding frequencies.

The given distribution being of the less than type, 140, 145, 150, 155, 160, 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150, 150-155, 155-160, 160-165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e. the frequency of class interval below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140 – 145 is 11 – 4 = 7. Similarly, the frequency of 145 – 150 is 29 – 11 = 18, for 150 – 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with given cumulative frequencies becomes:

Now n = 51. S0, n/ 2 = 51/ 2 =25.5 this observation lies in the class 145 – 150

Then,

L (the lower limit) = 145

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

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

h (the class size) = 5

Using the formula, median = l+ ×h, we have

Median = Missing close brace

= 145 + 72.5/ 18 = 149.03

So, the median height of the girls is 149.03 cm

This means that the height of about 50% of the girls in less than this height, and 50% are taller than this height.

17.A life insurance agent found the following data for 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.

Sol: Here class width is not same. There is no need to adjust the frequencies according to class interval. Now given frequencies table is less type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years we can define class intervals with their respective cumulative frequency as below.

Now from table we may observe that n = 100

Cumulative frequency (cf) just greater than n/ 2 (i.e. 100/ 2 = 50) is 78 belonging to interval 35 – 40

So median class                = 35 – 40

Lower limit (l) of median class = 35

Class size (h) = 5

Frequency (f) of median class = 33

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

Median = 

= 35+25/33

= 35.76

So 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:

 Find the mean length of life

Sol: The given data is not having continuous class intervals. We can observe the difference between 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 continuous class intervals with respective cumulative frequencies can be represented as below:

From the table we may observe that cumulative frequency just greater then n/ 2 (i.e. 40/ 2 =20) is 29, belongs to class interval 144.5 – 153.5

Median class = 144.5 – 153.5

Lower limit (l) = 144.5

Class size (h) = 9

Frequency (f) of median class = 12

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

Median = 

= 144.5 + 9/ 4 = 146.75

So median length of leaves is 146.75mm

19.An incomplete distribution is given as follows:

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

Sol:

Given

Median = 35

Then median class = 30 – 40

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

Median = 

20 = 55 – f₁

f₁ = 55 – 20 = 35

Given

Sum of frequencies = 170

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

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

f₂ = 25

f₁ = 35 and f₂ =25