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# Class 10 RD Sharma Solution – Chapter 7 Statistics – Exercise 7.4 | Set 1

• Last Updated : 03 Mar, 2021

### Question 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.

Solution:

On arranging the observations in ascending order, we have

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

Number of terms in the observation sequence is odd, i.e., N = 15

Now,

Median = (N + 1)/2th term

= (15 + 1)/2th term

= 8th term

Therefore, 716, which is the 8th term is the median of the data.

### Find the median height.

Solution:

We have N = 420,

So, N/2 = 420/ 2 = 210

Now, The cumulative frequency just greater than N/2 is 275

Therefore, 165.5 – 168.5 is the median class s.t

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3 = 165.5 + 1.63

= 167.13

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

Solution:

N = 100,

Therefore, N/2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) s.t,

L = 94.5, F = 33, h = (104.5 – 94.5) = 10 = 94.5 + 4.85

= 99.35

### Question 4. Calculate the median from the following data:

Solution:

N = 140,

And, N/2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 s.t,

L = 55, f = 40, F = 58, h = 65 – 55 = 10 ### Question 5. Calculate the median from the following data:

Solution:

N = 250,

And, N/2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 s.t,

L = 50, f = 31, F = 96, h = 60 -50 = 10 ### Question 6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Solution:

Let us assume the unknown frequency to be x.

Given: Median = 24

Therefore,

Median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30 4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, x = 25

### Calculate the median and interpret the results.

Solution:

N = 357,

And, N/2 = 357/ 2 = 178.5

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

Therefore, median class is 19.5 – 24.5 s.t

l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5 Median = 23.98, that implies that nearly half of the women are married between the ages of 15 and 25.

### Find the median life.

Solution:

Now

N = 400

And the cumulative frequency just greater than n/2 (= 200) is 216, which belongs to the class interval 3000 – 3500

Median class = 3000 – 3500. Therefore,

(l)  = 3000 and,(f) of median class = 86, (cf) of class preceding median class = 130 and (h) = 500

We have, = 3000 + (35000/86)

= 3406.98 hrs, which is the median time of lamps.

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

Solution:

The cf value just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.

Therefore,

Median class = 55 – 60

where,

(l) of median class = 55, (f) of median class = 6, (cf) = 13 and (h) = 5

We have, = 55 + 10/6 = 56.666 which is approximately 56.67 kg.

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

Solution:

Since, we know,

N = 200

Substituting values, we get,

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114…… (i)

Also, Mean = 1.46

⇒ Sum/ N = 1.46

Substituting values,

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 …….(ii)

Solving from (i) and (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

And, x = 114 – 38 = 76 (from equation (i))

Now, putting the values, we get,

N = 200 N/2 = 200/2 = 100

So, the cumulative frequency just greater than N/2 is 122

And, therefore, the median is 1.

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