Search

Samacheer Kalvi 10th Maths Unit Exercise 8 Solutions

10th Maths Unit Exercise 8 solutions (Chapter 8 Statistics and Probability) with step-by-step answers for all 11 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance11 answers · full steps below
  1. Q1f1=8,  f2=12f_1=8,\; f_2=12
  2. Q2Standard deviation ≈5.55\approx 5.55 mm
  3. Q3k=7k=7
  4. Q4Variance =81  (oF)2=81\;(^{o}F)^2
  5. Q5Mean =316≈5.17=\dfrac{31}{6}\approx 5.17, Standard deviation =856≈1.54=\dfrac{\sqrt{85}}{6}\approx 1.54
  6. Q6C.V. of city A =12.25%=12.25\%, C.V. of city B =24.59%=24.59\%; prices were more stable in city A
  7. Q7Largest value =60=60, Smallest value =40=40
  8. Q819\dfrac{1}{9}
  9. Q934\dfrac{3}{4}
  10. Q10Number of black balls =10=10
  11. Q11P(passing Tamil)=0.65P(\text{passing Tamil})=0.65

Q1. The mean of the following frequency distribution is 62.862.8 and the sum …

The mean of the following frequency distribution is 62.862.8 and the sum of all frequencies is 5050. Compute the missing frequencies f1f_1 and f2f_2.

Class Interval 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 5 f1f_1 10 f2f_2 7 8
Show steps (10)

  1. Let the midpoints be xi=10,30,50,70,90,110x_i = 10, 30, 50, 70, 90, 110.

  2. Sum of frequencies: 5+f1+10+f2+7+8=505+f_1+10+f_2+7+8=50

  3. f1+f2=20f_1+f_2=20 ... (1)

  4. ∑fixi=50+30f1+500+70f2+630+880=2060+30f1+70f2\sum f_ix_i = 50+30f_1+500+70f_2+630+880 = 2060+30f_1+70f_2

  5. Mean =∑fixi∑fi=62.8=\dfrac{\sum f_ix_i}{\sum f_i}=62.8, so ∑fixi=62.8×50=3140\sum f_ix_i = 62.8\times 50 = 3140

  6. 2060+30f1+70f2=31402060+30f_1+70f_2=3140

  7. 30f1+70f2=1080⇒3f1+7f2=10830f_1+70f_2=1080 \Rightarrow 3f_1+7f_2=108 ... (2)

  8. From (1), f1=20−f2f_1=20-f_2. Put in (2):

  9. 3(20−f2)+7f2=108⇒60+4f2=1083(20-f_2)+7f_2=108 \Rightarrow 60+4f_2=108

  10. 4f2=48⇒f2=124f_2=48 \Rightarrow f_2=12, hence f1=8f_1=8

Answer

f1=8,  f2=12f_1=8,\; f_2=12

Q2. The diameter of circles (in mm) drawn in a design are given …

The diameter of circles (in mm) drawn in a design are given below. Calculate the standard deviation.

Diameters 33-36 37-40 41-44 45-48 49-52
Number of circles 15 17 21 22 25
Show steps (9)

  1. The classes are inclusive, so take true boundaries 32.5−36.5,  36.5−40.5,…32.5-36.5,\;36.5-40.5,\dots with class size c=4c=4.

  2. Midpoints xx: 34.5,  38.5,  42.5,  46.5,  50.534.5,\;38.5,\;42.5,\;46.5,\;50.5

  3. Take A=42.5A=42.5 and d=x−A4d=\dfrac{x-A}{4}:   −2, −1, 0, 1, 2\;-2,\,-1,\,0,\,1,\,2

  4. ff: 15, 17, 21, 22, 2515,\,17,\,21,\,22,\,25, so N=∑f=100N=\sum f=100

  5. ∑fd=−30−17+0+22+50=25\sum fd = -30-17+0+22+50 = 25

  6. ∑fd2=60+17+0+22+100=199\sum fd^2 = 60+17+0+22+100 = 199

  7. σ=c∑fd2N−(∑fdN)2\sigma = c\sqrt{\dfrac{\sum fd^2}{N}-\left(\dfrac{\sum fd}{N}\right)^2}

  8. =4199100−(25100)2=41.9275= 4\sqrt{\dfrac{199}{100}-\left(\dfrac{25}{100}\right)^2} = 4\sqrt{1.9275}

  9. =4×1.3883=5.55= 4\times 1.3883 = 5.55

Answer

Standard deviation ≈5.55\approx 5.55 mm

Q3. The frequency distribution is given below. In the table, kk is a …

The frequency distribution is given below. In the table, kk is a positive integer, has a varience of 160. Determine the value of kk.

xx kk 2k2k 3k3k 4k4k 5k5k 6k6k
ff 2 1 1 1 1 1
Show steps (7)

  1. N=∑f=2+1+1+1+1+1=7N=\sum f = 2+1+1+1+1+1 = 7

  2. ∑fx=2k+2k+3k+4k+5k+6k=22k\sum fx = 2k+2k+3k+4k+5k+6k = 22k

  3. ∑fx2=2k2+4k2+9k2+16k2+25k2+36k2=92k2\sum fx^2 = 2k^2+4k^2+9k^2+16k^2+25k^2+36k^2 = 92k^2

  4. Variance =∑fx2N−(∑fxN)2=\dfrac{\sum fx^2}{N}-\left(\dfrac{\sum fx}{N}\right)^2

  5. =92k27−484k249=644k2−484k249=160k249=\dfrac{92k^2}{7}-\dfrac{484k^2}{49}=\dfrac{644k^2-484k^2}{49}=\dfrac{160k^2}{49}

  6. Given variance =160=160:

  7. 160k249=160⇒k2=49⇒k=7\dfrac{160k^2}{49}=160 \Rightarrow k^2=49 \Rightarrow k=7 (since kk is positive)

Answer

k=7k=7

Q4. The standard deviation of some temperature data in degree celsius (o^{o}C) is …

The standard deviation of some temperature data in degree celsius (o^{o}C) is 5. If the data were converted into degree Farenheit (o^{o}F) then what is the variance?

  1. Conversion: F=95C+32F=\dfrac{9}{5}C+32.

  2. Standard deviation is unchanged by adding 3232 but is multiplied by 95\dfrac{9}{5}.

  3. σF=95×5=9\sigma_F=\dfrac{9}{5}\times 5 = 9

  4. Variance =σF2=92=81=\sigma_F^2 = 9^2 = 81

Answer

Variance =81  (oF)2=81\;(^{o}F)^2

Q5. If for a distribution, ∑(x−5)=3\sum (x-5) = 3 , ∑(x−5)2=43\sum (x-5)^2 = 43 , and total number of …

If for a distribution, ∑(x−5)=3\sum (x-5) = 3, ∑(x−5)2=43\sum (x-5)^2 = 43, and total number of observations is 18, find the mean and standard deviation.

  1. Here d=x−5d=x-5, n=18n=18, ∑d=3\sum d=3, ∑d2=43\sum d^2=43.

  2. Mean =A+∑dn=5+318=5+16=316≈5.17=A+\dfrac{\sum d}{n}=5+\dfrac{3}{18}=5+\dfrac{1}{6}=\dfrac{31}{6}\approx 5.17

  3. σ=∑d2n−(∑dn)2\sigma=\sqrt{\dfrac{\sum d^2}{n}-\left(\dfrac{\sum d}{n}\right)^2}

  4. =4318−(16)2=86−136=8536=\sqrt{\dfrac{43}{18}-\left(\dfrac{1}{6}\right)^2}=\sqrt{\dfrac{86-1}{36}}=\sqrt{\dfrac{85}{36}}

  5. =856=9.21956≈1.54=\dfrac{\sqrt{85}}{6}=\dfrac{9.2195}{6}\approx 1.54

Answer

Mean =316≈5.17=\dfrac{31}{6}\approx 5.17, Standard deviation =856≈1.54=\dfrac{\sqrt{85}}{6}\approx 1.54

Q6. Prices of peanut packets in various places of two cities are given …

Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Prices in city A 20 22 19 23 16
Prices in city B 10 20 18 12 15
Show steps (9)

  1. City A: xˉ=20+22+19+23+165=1005=20\bar{x}=\dfrac{20+22+19+23+16}{5}=\dfrac{100}{5}=20

  2. Deviations: 0,2,−1,3,−40,2,-1,3,-4; ∑d2=0+4+1+9+16=30\sum d^2 = 0+4+1+9+16=30

  3. σA=305=6=2.449\sigma_A=\sqrt{\dfrac{30}{5}}=\sqrt{6}=2.449

  4. C.V.A=σxˉ×100=2.44920×100=12.25%_A=\dfrac{\sigma}{\bar{x}}\times 100=\dfrac{2.449}{20}\times100=12.25\%

  5. City B: xˉ=10+20+18+12+155=755=15\bar{x}=\dfrac{10+20+18+12+15}{5}=\dfrac{75}{5}=15

  6. Deviations: −5,5,3,−3,0-5,5,3,-3,0; ∑d2=25+25+9+9+0=68\sum d^2=25+25+9+9+0=68

  7. σB=685=13.6=3.688\sigma_B=\sqrt{\dfrac{68}{5}}=\sqrt{13.6}=3.688

  8. C.V.B=3.68815×100=24.59%_B=\dfrac{3.688}{15}\times100=24.59\%

  9. Since C.V.A<_A< C.V.B_B, city A is more consistent.

Answer

C.V. of city A =12.25%=12.25\%, C.V. of city B =24.59%=24.59\%; prices were more stable in city A

Q7. If the range and coefficient of range of the data are 2020 …

If the range and coefficient of range of the data are 2020 and 0.20.2 respectively, then find the largest and smallest values of the data.

  1. Range R=L−S=20R=L-S=20

  2. Coefficient of range =L−SL+S=0.2=\dfrac{L-S}{L+S}=0.2

  3. 20L+S=0.2⇒L+S=100\dfrac{20}{L+S}=0.2 \Rightarrow L+S=100

  4. Adding: 2L=120⇒L=602L=120 \Rightarrow L=60

  5. Then S=100−60=40S=100-60=40

Answer

Largest value =60=60, Smallest value =40=40

Q8. If two dice are rolled, then find the probability of getting the …

If two dice are rolled, then find the probability of getting the product of face value 66 or the difference of face values 55.

Show steps (8)

  1. Two dice are rolled, n(S)=36n(S)=36.

  2. AA = product of face values is 66

  3. A={(1,6),(6,1),(2,3),(3,2)}A=\{(1,6),(6,1),(2,3),(3,2)\}, n(A)=4n(A)=4

  4. BB = difference of face values is 55

  5. B={(1,6),(6,1)}B=\{(1,6),(6,1)\}, n(B)=2n(B)=2

  6. A∩B={(1,6),(6,1)}A\cap B=\{(1,6),(6,1)\}, n(A∩B)=2n(A\cap B)=2

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

  8. =436+236−236=436=19=\dfrac{4}{36}+\dfrac{2}{36}-\dfrac{2}{36}=\dfrac{4}{36}=\dfrac{1}{9}

Answer

19\dfrac{1}{9}

Q9. In a two children family, find the probability that there is at …

In a two children family, find the probability that there is at least one girl in a family.

  1. For a family with two children, S={BB, BG, GB, GG}S=\{BB,\,BG,\,GB,\,GG\}, n(S)=4n(S)=4.

  2. AA = at least one girl ={BG, GB, GG}=\{BG,\,GB,\,GG\}, n(A)=3n(A)=3

  3. P(A)=34P(A)=\dfrac{3}{4}

Answer

34\dfrac{3}{4}

Q10. A bag contains 55 white and some black balls. If the probability …

A bag contains 55 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.

  1. Let the number of black balls be xx. Total balls =5+x=5+x.

  2. P(white)=55+xP(\text{white})=\dfrac{5}{5+x}, P(black)=x5+xP(\text{black})=\dfrac{x}{5+x}

  3. Given P(black)=2 P(white)P(\text{black})=2\,P(\text{white})

  4. x5+x=2×55+x\dfrac{x}{5+x}=2\times\dfrac{5}{5+x}

  5. x=10x=10

Answer

Number of black balls =10=10

Q11. The probability that a student will pass the final examination in both …

The probability that a student will pass the final examination in both English and Tamil is 0.50.5 and the probability of passing neither is 0.10.1. If the probability of passing the English examination is 0.750.75, what is the probability of passing the Tamil examination?

Show steps (6)

  1. Let EE = passing English, TT = passing Tamil.

  2. P(E∩T)=0.5P(E\cap T)=0.5, P(E)=0.75P(E)=0.75

  3. P(E∪T‾)=0.1⇒P(E∪T)=1−0.1=0.9P(\overline{E\cup T})=0.1 \Rightarrow P(E\cup T)=1-0.1=0.9

  4. P(E∪T)=P(E)+P(T)−P(E∩T)P(E\cup T)=P(E)+P(T)-P(E\cap T)

  5. 0.9=0.75+P(T)−0.50.9=0.75+P(T)-0.5

  6. P(T)=0.9−0.25=0.65P(T)=0.9-0.25=0.65

Answer

P(passing Tamil)=0.65P(\text{passing Tamil})=0.65

Quick check

  1. 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 1 of 5

  2. 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. (1) 55
    2. (2) 1010
    3. (3) 1515
    4. (4) 2020
    Show answer

    Answer: (3) 1515

    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

    Question 2 of 5

  3. 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 3 of 5

  4. 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 4 of 5

  5. 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. (1) 310\dfrac{3}{10}
    2. (2) 710\dfrac{7}{10}
    3. (3) 39\dfrac{3}{9}
    4. (4) 79\dfrac{7}{9}
    Show answer

    Answer: (2) 710\dfrac{7}{10}

    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}

    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

Spotted a mistake? Email us at [email protected] and we will fix it.