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Samacheer Kalvi 10th Maths Exercise 8.2 Solutions

10th Maths Exercise 8.2 solutions (Chapter 8 Statistics and Probability) with step-by-step answers for all 8 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below
  1. Q1C.V. =52%= 52\%
  2. Q2Mean xˉ=4.6875\bar{x} = 4.6875 (approximately 4.694.69)
  3. Q3Standard deviation σ=7.2\sigma = 7.2
  4. Q4σ=117≈10.8\sigma=\sqrt{117}\approx 10.8 and C.V. ≈180.3%\approx 180.3\%
  5. Q5C.V. =14.4%= 14.4\%
  6. Q6C.V. =10.06%= 10.06\%
  7. Q7C.V. of Sathya =5%=5\%, C.V. of Vidhya =2.5%=2.5\%; Vidhya is more consistent in performance.
  8. Q8Social Science (16.67%16.67\%) is the most consistent and Science (21.54%21.54\%) is the least consistent.

Q1. The standard deviation and mean of a data are 6.56.5 and 12.512.5 …

The standard deviation and mean of a data are 6.56.5 and 12.512.5 respectively. Find the coefficient of variation.

  1. σ=6.5\sigma = 6.5, xˉ=12.5\bar{x} = 12.5

  2. C.V. =σxˉ×100%=\dfrac{\sigma}{\bar{x}}\times 100\%

  3. =6.512.5×100%=\dfrac{6.5}{12.5}\times 100\%

  4. =52%=52\%

Answer

C.V. =52%= 52\%

Q2. The standard deviation and coefficient of variation of a data are 1.21.2 …

The standard deviation and coefficient of variation of a data are 1.21.2 and 25.625.6 respectively. Find the value of mean.

  1. σ=1.2\sigma = 1.2, C.V. =25.6= 25.6

  2. C.V. =σxˉ×100%=\dfrac{\sigma}{\bar{x}}\times 100\%

  3. 25.6=1.2xˉ×10025.6=\dfrac{1.2}{\bar{x}}\times 100

  4. xˉ=1.2×10025.6=12025.6\bar{x}=\dfrac{1.2\times 100}{25.6}=\dfrac{120}{25.6}

  5. xˉ=4.6875\bar{x}=4.6875

Answer

Mean xˉ=4.6875\bar{x} = 4.6875 (approximately 4.694.69)

Q3. If the mean and coefficient of variation of a data are 1515 …

If the mean and coefficient of variation of a data are 1515 and 4848 respectively, then find the value of standard deviation.

  1. xˉ=15\bar{x} = 15, C.V. =48= 48

  2. C.V. =σxˉ×100%=\dfrac{\sigma}{\bar{x}}\times 100\%

  3. 48=σ15×10048=\dfrac{\sigma}{15}\times 100

  4. σ=48×15100\sigma=\dfrac{48\times 15}{100}

  5. σ=7.2\sigma=7.2

Answer

Standard deviation σ=7.2\sigma = 7.2

Q4. If n=5n = 5 , xˉ=6\bar{x} = 6 , Σx2=765\Sigma x^2 = 765 , then calculate the coefficient of …

If n=5n = 5, xˉ=6\bar{x} = 6, Σx2=765\Sigma x^2 = 765, then calculate the coefficient of variation.

  1. σ=Σx2n−(Σxn)2\sigma=\sqrt{\dfrac{\Sigma x^{2}}{n}-\left(\dfrac{\Sigma x}{n}\right)^{2}}

  2. =7655−62=\sqrt{\dfrac{765}{5}-6^{2}}

  3. =153−36=117=10.8=\sqrt{153-36}=\sqrt{117}=10.8 (approx.)

  4. C.V. =σxˉ×100%=10.86×100%=\dfrac{\sigma}{\bar{x}}\times 100\%=\dfrac{10.8}{6}\times 100\%

  5. =180.3%=180.3\% (approx.)

Answer

σ=117≈10.8\sigma=\sqrt{117}\approx 10.8 and C.V. ≈180.3%\approx 180.3\%

Q5. Find the coefficient of variation of 24,26,33,37,29,3124, 26, 33, 37, 29, 31 .

  1. xˉ=24+26+33+37+29+316=1806=30\bar{x}=\dfrac{24+26+33+37+29+31}{6}=\dfrac{180}{6}=30

  2. Deviations d=x−30d=x-30: −6, −4, 3, 7, −1, 1-6,\,-4,\,3,\,7,\,-1,\,1

  3. Σd2=36+16+9+49+1+1=112\Sigma d^{2}=36+16+9+49+1+1=112

  4. σ=Σd2n=1126=18.67=4.32\sigma=\sqrt{\dfrac{\Sigma d^{2}}{n}}=\sqrt{\dfrac{112}{6}}=\sqrt{18.67}=4.32

  5. C.V. =4.3230×100%=14.4%=\dfrac{4.32}{30}\times 100\%=14.4\%

Answer

C.V. =14.4%= 14.4\%

Q6. The time taken (in minutes) to complete a homework by 88 students …

The time taken (in minutes) to complete a homework by 88 students in a day are given by 38,40,47,44,46,43,49,5338, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.

  1. xˉ=38+40+47+44+46+43+49+538=3608=45\bar{x}=\dfrac{38+40+47+44+46+43+49+53}{8}=\dfrac{360}{8}=45

  2. Deviations d=x−45d=x-45: −7, −5, 2, −1, 1, −2, 4, 8-7,\,-5,\,2,\,-1,\,1,\,-2,\,4,\,8

  3. Σd2=49+25+4+1+1+4+16+64=164\Sigma d^{2}=49+25+4+1+1+4+16+64=164

  4. σ=1648=20.5=4.53\sigma=\sqrt{\dfrac{164}{8}}=\sqrt{20.5}=4.53

  5. C.V. =4.5345×100%=10.06%=\dfrac{4.53}{45}\times 100\%=10.06\%

Answer

C.V. =10.06%= 10.06\%

Q7. The total marks scored by two students Sathya and Vidhya in 55 …

The total marks scored by two students Sathya and Vidhya in 55 subjects are 460460 and 480480 with standard deviation 4.64.6 and 2.42.4 respectively. Who is more consistent in performance?

  1. Sathya: xˉ1=4605=92\bar{x}_1=\dfrac{460}{5}=92, σ1=4.6\sigma_1=4.6

  2. C.V.1=4.692×100%=5%_1=\dfrac{4.6}{92}\times 100\%=5\%

  3. Vidhya: xˉ2=4805=96\bar{x}_2=\dfrac{480}{5}=96, σ2=2.4\sigma_2=2.4

  4. C.V.2=2.496×100%=2.5%_2=\dfrac{2.4}{96}\times 100\%=2.5\%

  5. C.V.2<_2< C.V.1_1, so Vidhya's marks are less varying.

Answer

C.V. of Sathya =5%=5\%, C.V. of Vidhya =2.5%=2.5\%; Vidhya is more consistent in performance.

Q8. The mean and standard deviation of marks obtained by 4040 students of …

The mean and standard deviation of marks obtained by 4040 students of a class in three subjects Mathematics, Science and Social Science are given below. Which of the three subjects shows more consistent and which shows less consistent in marks?

Subject Mean SD
Mathematics 56 12
Science 65 14
Social Science 60 10

  1. Mathematics: C.V. =1256×100%=21.43%=\dfrac{12}{56}\times 100\%=21.43\%

  2. Science: C.V. =1465×100%=21.54%=\dfrac{14}{65}\times 100\%=21.54\%

  3. Social Science: C.V. =1060×100%=16.67%=\dfrac{10}{60}\times 100\%=16.67\%

  4. Least C.V. is for Social Science, greatest C.V. is for Science.

Answer

Social Science (16.67%16.67\%) is the most consistent and Science (21.54%21.54\%) is the least consistent.

Quick check

  1. The probability of getting a job for a person is x3\dfrac{x}{3}. If the probability of not getting the job is 23\dfrac{2}{3} then the value of xx is

    1. (1) 22
    2. (2) 11
    3. (3) 33
    4. (4) 1.51.5
    Show answer

    Answer: (2) 11

    1. P(A)+P(Aˉ)=1P(A)+P(\bar{A})=1

    2. x3+23=1\dfrac{x}{3}+\dfrac{2}{3}=1

    3. x3=13⇒x=1\dfrac{x}{3}=\dfrac{1}{3}\Rightarrow x=1

    Question 1 of 5

  2. The mean of 100100 observations is 4040 and their standard deviation is 33. The sum of squares of all observations is

    1. (1) 4000040000
    2. (2) 160900160900
    3. (3) 160000160000
    4. (4) 3000030000
    Show answer

    Answer: (2) 160900160900

    1. σ2=∑x2n−(xˉ)2\sigma^2=\dfrac{\sum x^2}{n}-\left(\bar{x}\right)^2

    2. 9=∑x2100−16009=\dfrac{\sum x^2}{100}-1600

    3. ∑x2100=1609\dfrac{\sum x^2}{100}=1609

    4. ∑x2=160900\sum x^2=160900

    Question 2 of 5

  3. Which of the following is incorrect?

    1. (1) P(A)>1P(A) > 1
    2. (2) 0≤P(A)≤10 \leq P(A) \leq 1
    3. (3) P(ϕ)=0P(\phi) = 0
    4. (4) P(A)+P(Aˉ)=1P(A) + P(\bar{A}) = 1
    Show answer

    Answer: (1) P(A)>1P(A) > 1

    1. Probability of any event always lies between 00 and 11.

    2. So P(A)>1P(A)>1 is impossible.

    Question 3 of 5

  4. The probability a red marble selected at random from a jar containing pp red, qq blue and r green marbles is

    1. (1) qp+q+r\dfrac{q}{p+q+r}
    2. (2) pp+q+r\dfrac{p}{p+q+r}
    3. (3) p+qp+q+r\dfrac{p+q}{p+q+r}
    4. (4) p+rp+q+r\dfrac{p+r}{p+q+r}
    Show answer

    Answer: (2) pp+q+r\dfrac{p}{p+q+r}

    1. Total marbles =p+q+r=p+q+r; red marbles =p=p.

    2. P(red)=pp+q+rP(\text{red})=\dfrac{p}{p+q+r}

    Question 4 of 5

  5. The range of the data 8,8,8,8,8...88, 8, 8, 8, 8. . . 8 is

    1. (1) 00
    2. (2) 11
    3. (3) 88
    4. (4) 33
    Show answer

    Answer: (1) 00

    1. Largest value =8= 8, smallest value =8= 8.

    2. Range =L−S=8−8=0= L - S = 8 - 8 = 0.

    Question 5 of 5
Chapter 8 formulas
  • Range

    R=L−S(L - largest value, S - smallest value)R = L - S \quad (L\text{ - largest value},\ S\text{ - smallest value})

  • Coefficient of range

    Coefficient of range=L−SL+S\text{Coefficient of range} = \frac{L-S}{L+S}

  • Variance

    σ2=∑i=1n(xi−xˉ)2n\sigma^{2} = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^{2}}{n}

  • Standard deviation

    σ=∑(xi−xˉ)2n\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^{2}}{n}}

  • Standard deviation (ungrouped data) - Direct method

    σ=∑xi2n−(∑xin)2\sigma = \sqrt{\frac{\sum x_i^{2}}{n} - \left(\frac{\sum x_i}{n}\right)^{2}}

  • Standard deviation (ungrouped data) - Mean method

    σ=∑di2n\sigma = \sqrt{\frac{\sum d_i^{2}}{n}}

  • Standard deviation (ungrouped data) - Assumed mean method

    σ=∑di2n−(∑din)2\sigma = \sqrt{\frac{\sum d_i^{2}}{n} - \left(\frac{\sum d_i}{n}\right)^{2}}

  • Standard deviation (ungrouped data) - Step deviation method

    σ=c×∑di2n−(∑din)2\sigma = c \times \sqrt{\frac{\sum d_i^{2}}{n} - \left(\frac{\sum d_i}{n}\right)^{2}}

  • Standard deviation of first n natural numbers

    σ=n2−112\sigma = \sqrt{\frac{n^{2}-1}{12}}

  • Standard deviation (grouped data) - Mean method

    σ=∑fidi2N\sigma = \sqrt{\frac{\sum f_i d_i^{2}}{N}}

  • Standard deviation (grouped data) - Assumed mean method

    σ=∑fidi2N−(∑fidiN)2\sigma = \sqrt{\frac{\sum f_i d_i^{2}}{N} - \left(\frac{\sum f_i d_i}{N}\right)^{2}}

  • Standard deviation (grouped data) - Step deviation method

    σ=C×∑fidi2N−(∑fidiN)2\sigma = C \times \sqrt{\frac{\sum f_i d_i^{2}}{N} - \left(\frac{\sum f_i d_i}{N}\right)^{2}}

  • Coefficient of variation

    C.V=σxˉ×100%\text{C.V} = \frac{\sigma}{\bar{x}} \times 100\%

  • Consistency using C.V.

    Smaller C.V.⇒more consistent data;larger C.V.⇒less consistent data\text{Smaller C.V.} \Rightarrow \text{more consistent data};\quad \text{larger C.V.} \Rightarrow \text{less consistent data}

  • Mutually exclusive events

    A, B are mutually exclusive if A∩B=ϕA,\ B \text{ are mutually exclusive if } A \cap B = \phi

  • Probability of an event

    P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}

  • Sure and impossible events

    P(S)=1,P(ϕ)=0P(S) = 1,\qquad P(\phi) = 0

  • Range of probability

    0≤P(E)≤10 \le P(E) \le 1

  • Complement of an event

    P(Eˉ)=1−P(E)P(\bar{E}) = 1 - P(E)

  • Addition theorem (mutually exclusive events)

    P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

  • Probability of only A

    P(A∩Bˉ)=P(only A)=P(A)−P(A∩B)P(A \cap \bar{B}) = P(\text{only } A) = P(A) - P(A \cap B)

  • Probability of only B

    P(Aˉ∩B)=P(only B)=P(B)−P(A∩B)P(\bar{A} \cap B) = P(\text{only } B) = P(B) - P(A \cap B)

  • Addition theorem for any two events

    P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

  • Addition theorem for any three events

    P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(C∩A)+P(A∩B∩C)P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)

Answers checked against the Samacheer Kalvi textbook · Updated

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