📊 Grade 10 Maths · Unit 8 · Samacheer Kalvi

Samacheer Class 10 Maths - Statistics and Probability

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Complete step-by-step solutions for every exercise in Unit 8. Click any question to expand the full working.

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📋 Exercises in this unit
Ex 8.1 — Measures of Dispersion Ex 8.1 — --
Your Progress — Unit 8: Statistics and Probability 0% complete
Ex 8.1Measures of Dispersion1 questions
Q.1 Question 1
✓ Solution

Mathematics : Statistics And Probability : Measures of Dispersion : Exercise Questions with Answers

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Ex 8.1--15 questions
Q.1 Find the range and coefficient of range of the following data.
✓ Solution
(i) 63, 89, 98, 125, 79, 108, 117, 68

Largest value:

$$ L = 125 $$

Smallest value:

$$ S = 63 $$

Range:

$$ = L-S $$

$$ =125-63 $$

$$ =62 $$

Coefficient of range:

$$ =\frac{L-S}{L+S} $$

$$ =\frac{125-63}{125+63} $$

$$ =\frac{62}{188} $$

$$ \approx0.33 $$

### Answer

Range:

$$ 62 $$

Coefficient of range:

$$ 0.33 $$

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(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8

Largest value:

$$ 61.4 $$

Smallest value:

$$ 13.6 $$

Range:

$$ 61.4-13.6 $$

$$ =47.8 $$

Coefficient of range:

$$ =\frac{47.8}{61.4+13.6} $$

$$ =\frac{47.8}{75} $$

$$ \approx0.64 $$

### Answer

Range:

$$ 47.8 $$

Coefficient of range:

$$ 0.64 $$

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Q.2 If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
✓ Solution

Range:

$$ =L-S $$

Given:

$$ 36.8=L-13.4 $$

$$ L=36.8+13.4 $$

$$ L=50.2 $$

### Answer

$$ 50.2 $$

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Q.3 Calculate the range of the given data.
✓ Solution

Range:

$$ =L-S $$

After identifying the largest and smallest observations from the table,

$$ \text{Range}=250 $$

### Answer

$$ 250 $$

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Q.4 A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages.
✓ Solution

Pages yet to complete:

$$ 28,25,23,30,27,24,25,23 $$

Mean:

$$ \bar{x}=\frac{28+25+23+30+27+24+25+23}{8} $$

$$ =\frac{205}{8} $$

$$ =25.625 $$

Using

$$ \sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{n}} $$

After calculation,

$$ \sigma \approx 2.34 $$

### Answer

$$ 2.34 $$

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Q.5 Find the variance and standard deviation of the wages of 9 workers given below:
✓ Solution

Mean:

$$ \bar{x}=300 $$

Using variance formula,

$$ \sigma^2=\frac{\sum(x-\bar{x})^2}{n} $$

$$ =222.22 $$

Standard deviation:

$$ \sigma=\sqrt{222.22} $$

$$ \approx14.91 $$

### Answer

Variance:

$$ 222.22 $$

Standard deviation:

$$ 14.91 $$

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Q.6 A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock and so on.
✓ Solution

Total strikes in 12 hours:

$$ 1+2+3+\cdots+12 $$

$$ =\frac{12(13)}{2} $$

$$ =78 $$

In one day:

$$ 2\times78=156 $$

Standard deviation:

$$ 6.9 $$

### Answer

Total strikes:

$$ 156 $$

Standard deviation:

$$ 6.9 $$

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Q.7 Find the standard deviation of first 21 natural numbers.
✓ Solution

Natural numbers:

$$ 1,2,3,\ldots,21 $$

Mean:

$$ \bar{x}=\frac{21+1}{2}=11 $$

Using standard deviation formula,

$$ \sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{n}} $$

$$ \sigma\approx6.05 $$

### Answer

$$ 6.05 $$

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Q.8 If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
✓ Solution

Subtracting a constant from every observation does not change standard deviation.

### Answer

$$ 4.5 $$

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Q.9 If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
✓ Solution

New standard deviation:

$$ =\frac{3.6}{3} $$

$$ =1.2 $$

New variance:

$$ =(1.2)^2 $$

$$ =1.44 $$

### Answer

Variance:

$$ 1.44 $$

Standard deviation:

$$ 1.2 $$

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Q.10 The rainfall recorded in various places of five districts in a week are given below.
✓ Solution

$$ 7.76 $$

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Q.11 In a study about viral fever, the number of people affected in a town were noted.
✓ Solution

$$ 14.6 $$

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Q.12 The measurements of the diameters (in cms) of the plates prepared in a factory are given below.
✓ Solution

$$ 6 $$

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Q.13 The time taken by 50 students to complete a 100 meter race are given below.
✓ Solution

$$ 1.24 $$

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Q.14 For a group of 100 candidates the mean and standard deviation of their marks were found to be 60 and 15 respectively.
✓ Solution

Corrected mean:

$$ 60.5 $$

Corrected standard deviation:

$$ 14.61 $$

### Answer

Mean:

$$ 60.5 $$

Standard deviation:

$$ 14.61 $$

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Q.15 The mean and variance of seven observations are 8 and 16 respectively.
✓ Solution

Let remaining observations be $x$ and $y$.

Mean:

$$ \frac{2+4+10+12+14+x+y}{7}=8 $$

$$ 42+x+y=56 $$

$$ x+y=14 $$

Using variance formula:

$$ \sigma^2=\frac{\sum x^2}{n}-\bar{x}^2 $$

$$ 16=\frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7}-64 $$

$$ x^2+y^2=100 $$

Using:

$$ (x+y)^2=x^2+y^2+2xy $$

$$ 196=100+2xy $$

$$ xy=48 $$

Thus,

$$ t^2-14t+48=0 $$

$$ (t-6)(t-8)=0 $$

Hence,

$$ t=6,8 $$

### Answer

$$ 6 \text{ and } 8 $$

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# Answer Key

| Q.No | Answer | |---|---| | 1(i) | 62 ; 0.33 | | 1(ii) | 47.8 ; 0.64 | | 2 | 50.2 | | 3 | 250 | | 4 | 2.34 | | 5 | 222.22 ; 14.91 | | 6 | 6.9 | | 7 | 6.05 | | 8 | 4.5 | | 9 | 1.44 ; 1.2 | | 10 | 7.76 | | 11 | 14.6 | | 12 | 6 | | 13 | 1.24 | | 14 | 60.5 ; 14.61 | | 15 | 6 and 8 |


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