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Samacheer Kalvi 10th Maths Exercise 8.5 Solutions

10th Maths Exercise 8.5 solutions (Chapter 8 Statistics and Probability) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1(C) Arithmetic mean
  2. Q2(A) 00
  3. Q3(C) zero
  4. Q4(B) 160900160900
  5. Q5(C) 33.2533.25
  6. Q6(D) 225225
  7. Q7(B) 3p3p
  8. Q8(A) 3.53.5
  9. Q9(A) P(A)>1P(A) > 1
  10. Q10(B) pp+q+r\dfrac{p}{p+q+r}
  11. Q11(B) 710\dfrac{7}{10}
  12. Q12(B) 11
  13. Q13(C) 1515
  14. Q14(C) 2326\dfrac{23}{26}
  15. Q15(D) 45\dfrac{4}{5}

Q1. Which of the following is not a measure of dispersion?

  1. (A) Range
  2. (B) Standard deviation
  3. (C) Arithmetic mean
  4. (D) Variance

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

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

Answer

(C) Arithmetic mean

Q2. The range of the data 8,8,8,8,8...88, 8, 8, 8, 8. . . 8 is

  1. (A) 00
  2. (B) 11
  3. (C) 88
  4. (D) 33

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

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

Answer

(A) 00

Q3. The sum of all deviations of the data from its mean is

  1. (A) Always positive
  2. (B) always negative
  3. (C) zero
  4. (D) non-zero integer

  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.

Answer

(C) zero

Q4. The mean of 100100 observations is 4040 and their standard deviation is …

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

  1. (A) 4000040000
  2. (B) 160900160900
  3. (C) 160000160000
  4. (D) 3000030000

  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

Answer

(B) 160900160900

Q5. Variance of first 2020 natural numbers is

  1. (A) 32.2532.25
  2. (B) 44.2544.25
  3. (C) 33.2533.25
  4. (D) 3030

  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

Answer

(C) 33.2533.25

Q6. The standard deviation of a data is 33 . If each value …

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

  1. (A) 33
  2. (B) 1515
  3. (C) 55
  4. (D) 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.

Answer

(D) 225225

Q7. If the standard deviation of x, y, zx,\ y,\ z is pp then the standard deviation …

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. (A) 3p+53p+5
  2. (B) 3p3p
  3. (C) p+5p+5
  4. (D) 9p+159p+15

  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.

Answer

(B) 3p3p

Q8. If the mean and coefficient of variation of a data are 44 …

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

  1. (A) 3.53.5
  2. (B) 33
  3. (C) 4.54.5
  4. (D) 2.52.5

  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

Answer

(A) 3.53.5

Q9. Which of the following is incorrect?

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

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

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

Answer

(A) P(A)>1P(A) > 1

Q10. The probability a red marble selected at random from a jar containing …

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

  1. (A) qp+q+r\dfrac{q}{p+q+r}
  2. (B) pp+q+r\dfrac{p}{p+q+r}
  3. (C) p+qp+q+r\dfrac{p+q}{p+q+r}
  4. (D) p+rp+q+r\dfrac{p+r}{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}

Answer

(B) pp+q+r\dfrac{p}{p+q+r}

Q11. A page is selected at random from a book. The probability that …

A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 77 is

  1. (A) 310\dfrac{3}{10}
  2. (B) 710\dfrac{7}{10}
  3. (C) 39\dfrac{3}{9}
  4. (D) 79\dfrac{7}{9}

  1. Units digit can be 0,1,2,…,90,1,2,\ldots,9, so n(S)=10n(S)=10.

  2. Digits less than 77: 0,1,2,3,4,5,60,1,2,3,4,5,6, so n(A)=7n(A)=7.

  3. P(A)=710P(A)=\dfrac{7}{10}

Answer

(B) 710\dfrac{7}{10}

Q12. The probability of getting a job for a person is x3\dfrac{x}{3} . …

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. (A) 22
  2. (B) 11
  3. (C) 33
  4. (D) 1.51.5

  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

Answer

(B) 11

Q13. Kamalam went to play a lucky draw contest. 135135 tickets of the …

Kamalam went to play a lucky draw contest. 135135 tickets of the lucky draw were sold. If the probability of Kamalam winning is 19\dfrac{1}{9}, then the number of tickets bought by Kamalam is

  1. (A) 55
  2. (B) 1010
  3. (C) 1515
  4. (D) 2020

  1. Let the number of tickets bought be nn.

  2. n135=19\dfrac{n}{135}=\dfrac{1}{9}

  3. n=1359=15n=\dfrac{135}{9}=15

Answer

(C) 1515

Q14. If a letter is chosen at random from the English alphabets {a,b,…,z}\{a,b,\ldots,z\} …

If a letter is chosen at random from the English alphabets {a,b,…,z}\{a,b,\ldots,z\}, then the probability that the letter chosen precedes xx

  1. (A) 1213\dfrac{12}{13}
  2. (B) 113\dfrac{1}{13}
  3. (C) 2326\dfrac{23}{26}
  4. (D) 326\dfrac{3}{26}

  1. Total letters n(S)=26n(S)=26.

  2. Letters before xx are aa to ww, that is 2323 letters.

  3. P=2326P=\dfrac{23}{26}

Answer

(C) 2326\dfrac{23}{26}

Q15. A purse contains 10 notes of ₹20002000, 1515 notes of ₹500500, and …

A purse contains 10 notes of ₹20002000, 1515 notes of ₹500500, and 2525 notes of ₹200200. One note is drawn at random. What is the probability that the note is either a ₹500500 note or ₹200200 note?

  1. (A) 15\dfrac{1}{5}
  2. (B) 310\dfrac{3}{10}
  3. (C) 23\dfrac{2}{3}
  4. (D) 45\dfrac{4}{5}

  1. Total notes =10+15+25=50=10+15+25=50.

  2. Favourable notes =15+25=40=15+25=40.

  3. P=4050=45P=\dfrac{40}{50}=\dfrac{4}{5}

Answer

(D) 45\dfrac{4}{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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