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10th Maths Chapter 8 Statistics and Probability

10th Maths Chapter 8 Statistics and Probability solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.

This chapter joins two topics. The first part measures how spread out data is, using range, coefficient of range, variance and standard deviation for both ungrouped and grouped values, and compares two sets through the coefficient of variation. The second part studies random experiments, sample space and events, and builds up the addition rules of probability. Exercises ask you to compute spread, test consistency and find probabilities for dice, cards, balls and similar situations.

Exercises in this chapter

Key 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)

Chapter 8 one-mark practice

  1. Which of the following is not a measure of dispersion?

    1. (1) Range
    2. (2) Standard deviation
    3. (3) Arithmetic mean
    4. (4) Variance
    Show answer

    Answer: (3) Arithmetic mean

    1. Dispersion measures how spread out the data is: range, standard deviation, variance.

    2. Arithmetic mean is a measure of central tendency, not dispersion.

    Question 1 of 10

  2. 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 2 of 10

  3. The sum of all deviations of the data from its mean is

    1. (1) Always positive
    2. (2) always negative
    3. (3) zero
    4. (4) non-zero integer
    Show answer

    Answer: (3) zero

    1. ∑(xi−xˉ)=∑xi−nxˉ\sum (x_i-\bar{x}) = \sum x_i - n\bar{x}.

    2. Since xˉ=∑xin\bar{x}=\dfrac{\sum x_i}{n}, we get nxˉ=∑xin\bar{x}=\sum x_i.

    3. So the sum of deviations =0=0.

    Question 3 of 10

  4. 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 4 of 10

  5. Variance of first 2020 natural numbers is

    1. (1) 32.2532.25
    2. (2) 44.2544.25
    3. (3) 33.2533.25
    4. (4) 3030
    Show answer

    Answer: (3) 33.2533.25

    1. Variance of first nn natural numbers =n2−112=\dfrac{n^2-1}{12}.

    2. =202−112=39912=33.25=\dfrac{20^2-1}{12}=\dfrac{399}{12}=33.25

    Question 5 of 10

  6. The standard deviation of a data is 33. If each value is multiplied by 55 then the new variance is

    1. (1) 33
    2. (2) 1515
    3. (3) 55
    4. (4) 225225
    Show answer

    Answer: (4) 225225

    1. Multiplying each value by 55 multiplies the S.D. by 55.

    2. New S.D. =5×3=15=5\times 3=15.

    3. New variance =152=225=15^2=225.

    Question 6 of 10

  7. If the standard deviation of x, y, zx,\ y,\ z is pp then the standard deviation of 3x+53x+5, 3y+53y+5, 3z+53z+5 is

    1. (1) 3p+53p+5
    2. (2) 3p3p
    3. (3) p+5p+5
    4. (4) 9p+159p+15
    Show answer

    Answer: (2) 3p3p

    1. Adding a constant does not change the S.D.; multiplying by 33 multiplies the S.D. by 33.

    2. New standard deviation =3p=3p.

    Question 7 of 10

  8. If the mean and coefficient of variation of a data are 44 and 87.5%87.5\% then the standard deviation is

    1. (1) 3.53.5
    2. (2) 33
    3. (3) 4.54.5
    4. (4) 2.52.5
    Show answer

    Answer: (1) 3.53.5

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

    2. 87.5=σ4×10087.5=\dfrac{\sigma}{4}\times 100

    3. σ=87.5×4100=3.5\sigma=\dfrac{87.5\times 4}{100}=3.5

    Question 8 of 10

  9. 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 9 of 10

  10. 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 10 of 10

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