{"id":12033,"date":"2019-12-13T15:15:49","date_gmt":"2019-12-13T09:45:49","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=12033"},"modified":"2021-07-02T15:36:48","modified_gmt":"2021-07-02T10:06:48","slug":"1st-puc-statistics-question-bank-chapter-10","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-statistics-question-bank-chapter-10\/","title":{"rendered":"1st PUC Statistics Question Bank Chapter 10 Random Variables and Mathematical Expectation"},"content":{"rendered":"

You can Download Chapter 10 Random Variables and Mathematical Expectation Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Statistics Question Bank Chapter 10 Random Variables and Mathematical Expectation<\/h2>\n

1st PUC Random Variables and Mathematical Expectation Two Marks Questions and Answers<\/h3>\n

Question 1.
\nDefine Random variable.
\nAnswer:
\nRandom variable is a function which assigns a real number to every sample point in the sample space.<\/p>\n

Question 2.
\nDefine discrete random variable
\nAnswer:
\nA random variable \u2018 X \u2019 which takes the specified values x1<\/sub>x2<\/sub>,…,xn<\/sub> with a respective probabilities p1<\/sub>, p2<\/sub>,….., p1<\/sub> is a discrete random variable.<\/p>\n

Question 3.
\nDefine continuous random variable.
\nAnswer:
\nA random variable which assumes all the possible values in its range is called a continuous random variable.<\/p>\n

Question 4.
\nDefine Mathematical expectation of a random variable\/Mean of random variable.
\nAnswer:
\nLet \u2018X\u2019 be a discrete random variable which can takes the values x1<\/sub>, x2<\/sub>, x3<\/sub>,… ,xn<\/sub> with respective probabilities p1<\/sub>, p2<\/sub>,p3<\/sub>,… , pn<\/sub> then, the mathematical expectation of \u2018 X \u2019 is:E(X) = x1<\/sub>p1<\/sub> + x2<\/sub>P2<\/sub> + x3<\/sub>P3<\/sub> + – + xn<\/sub>Pn<\/sub>
\nThen E(X) = \u03a3 x p(x) ;it is also called as mean of discrete random variable (X)<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nDefine probability distribution.
\nAnswer:
\nA systematic presentation of the values taken by a random variable with respective probabilities is called the probability distribution of a random variable\u2019.<\/p>\n

Question 6.
\nProbability mass function (pmf).
\nAnswer:
\nProbability mass function (pmf): Let \u2018 X \u2019 be a discrete random variable . And let p(x) be a function : p(x) = p(X=x) . Then p(x) is called the probability mass function of \u2018 X \u2019. If the following conditions are satisfied,<\/p>\n

    \n
  • p(x) \u2265 0 for all values X, and<\/li>\n
  • \u03a3 p(x) =1.<\/li>\n<\/ul>\n

    Question 7.
    \nProbability density function (pdf).
    \nAnswer:
    \nProbability density function (pdf): Let \u2018 X \u2019 be a continuous random variable taking values in the interval [a, b], then,
    \nA function f(x) is said to be the probability density function of the continuous random variable \u2018 X \u2019, if it satisfies the following conditions :<\/p>\n

      \n
    • f(x) \u2265 0 for all \u2018 X\u2019 in the interval [a, b] .<\/li>\n
    • For two distinct numbers c & d in the interval [a, b]:
      \nP(c \u2264 X \u2264 d) = (Area under the probability curve between ordinates at X =c and X=d).<\/li>\n
    • Total area under the curve is 1. i.e. , P(- \u221e < x < \u221e) = 1 .<\/li>\n<\/ul>\n

      Question 8.
      \nWritedown the formulae of all Mathematical Expectations:
      \nAnswer:<\/p>\n

        \n
      • E(X) = X\u03a3X.P (X)<\/li>\n
      • E(a) = a,<\/li>\n
      • E(aX) = a E(X) and<\/li>\n
      • E(aX+b) = aE(X)+b.<\/li>\n<\/ul>\n

        Question 9.
        \nWrite down the Formulae of Variances:
        \nAnswer:<\/p>\n

          \n
        • Var(X)= {E(X2<\/sup>)-[E(X)]2<\/sup>} ,<\/li>\n
        • V (a)=0,<\/li>\n
        • V(aX)=a2<\/sup> V(X),<\/li>\n
        • V(aX+b)=a2<\/sup>V(X), and S.D.(X)= \\(\\sqrt{\\mathrm{v}(\\mathrm{x})}\\).<\/li>\n<\/ul>\n

          Question 10.
          \nWrite down the Formula of Co-Variance:
          \nAnswer:
          \nCov(X, Y) =E(XY) – E(X).E(Y),<\/p>\n

          Question 11.
          \nWrite down the co-efficient of correlation:
          \nAnswer:
          \n\"1st<\/p>\n

          Question 12.
          \nState Addition theorem expectation
          \nAnswer:
          \nStatement: Let X and Y be two random variables with respective expectations E(X) and E(Y). Then, E(X + Y)=E(X) + E(Y)<\/p>\n

          Question 13.
          \nState Multiplication of expectation.
          \nAnswer:
          \nstatement: Let X and Y be two independent random variables with respective expectations E(X) and E(Y). Then expectation of the product of these random variables is; E(XY) = E(X) E(Y)<\/p>\n

          Question 14.
          \nIf E(X)=2, then find the value of E(-x\/4)
          \nAnswer:
          \nGiven E(-x\/4) = E(-l\/4. x) =-l\/4.E(X) = -1\/4. 2 =-1\/2 since E(aX) = a E(X)<\/p>\n

          Question 15.
          \nIf E(X)=2\/5, find E(5x\/3)
          \nAnswer:
          \nGiven E(5x\/3) = E(5\/3.x) = 5\/3.E(x) = 5\/3.2\/5 = 2\/3<\/p>\n

          \"KSEEB<\/p>\n

          Question 16.
          \nIf X is a random variable and E(X) =1.5, then what is the value of E(3 + 4X)?
          \nAnswer:
          \nE(3 + 4X) = E(4x + 3) = 4.E(x) + 3 = 4.(1.5) + 3 = 6 +3 = 9 since
          \nE(ax+b) = aE(x)+b<\/p>\n

          Question 17.
          \nIf E(X) = 2, find the value of E(3X – 6).
          \nAnswer:
          \nE(3X – 6) = 3E(x) – 6 = 3(2) – 6 =6 – 6 =0<\/p>\n

          Question 18.
          \nIf E(X+Y) =7 and E(X) =4, then find the value of E(Y).
          \nAnswer:
          \nWe know that E(X+Y) = E(X) + E(Y) ;
          \n7 = 4 + E(y)
          \nE(y) = 7 – 4 = 3<\/p>\n

          Question 19.
          \nIf E(X) = 3 and E(Y) = 5, then find E(3X+2Y), E(3X -Y)
          \nAnswer:
          \nE(3X + 2Y) = 3 E(x) + 2 E(y) = 3(3) + 2(5) = 9 + 10=19
          \nE(3X – Y) = 3 E(X) + (-1) E(Y) = 3(3)- 5 = 9 – 5 = 4<\/p>\n

          Question 20.
          \nIf X and Y are two independent random variable, E(XY)=10 and E(X)= 4, then what is the value of E(Y)?
          \nAnswer:
          \nWe know for any two independent random variables:
          \nE(XY) = E(X)E(Y);
          \n10 = 4 E(Y); E(Y)= 10\/4 = 2.5<\/p>\n

          Question 21.
          \nWhat is the value of co-efficient of correlation for two independent variables?
          \nAnswer:
          \nZero ie., r = 0<\/p>\n

          Question 22.
          \nFor a random variable X, E(X2<\/sup>) = 50 and V(X)=14, then what is E(X)?
          \nAnswer:
          \nWe know that V(X) = E(X2<\/sup>)-[E(x)]2<\/sup>
          \n14 = 50-[E(x)]2<\/sup>
          \n[E(x)]2<\/sup> = 50 – 14 = 36
          \nSquaring on both sides, we get E(x) = \\(\\sqrt{36}\\) = 6<\/p>\n

          Question 23.
          \nIf E(X)=3 and E(X2<\/sup>)=45, find the value of S.D.
          \nAnswer:
          \nWe know S.D(x)
          \n\\(\\begin{array}{l}{=\\sqrt{\\mathrm{V}(\\mathrm{x})}=\\sqrt{\\mathrm{E}(\\mathrm{X} 2)-[\\mathrm{E}(\\mathrm{x})] 2}} \\\\ {=\\sqrt{45-[3]^{2}=\\sqrt{45-9}}=\\sqrt{36}=6}\\end{array}\\)<\/p>\n

          Question 24.
          \nIf E(X) = 4, E(X2<\/sup>) = 25, find V(5X – 9).
          \nAnswer:
          \nV(5x-9)this is of the type V(aX+b)=a2<\/sup>V(X)
          \nV(5x-9) = 52<\/sup>. V(x)
          \nHere V(X) = E(X2<\/sup>) – [E(x)]2<\/sup>;
          \n= 25 – 42<\/sup> = 25 – 16 = 9
          \n\u2234 V(5\u00d7-9) = 52<\/sup>. V(x) = 25(9) = 225<\/p>\n

          Question 25.
          \nIf V(X)=4 then find V(2X)
          \nAnswer:
          \nV(2x) = 22<\/sup>V(x) = 4 (4) = 16; Since V(aX) = a2<\/sup> V(X)<\/p>\n

          \"KSEEB<\/p>\n

          Question 26.
          \nIf V(X) = 5 then find V(-5X).
          \nAnswer:
          \nV(-5X) = (5)2 <\/sup>V(x) = 25(5) = 125; Since V(aX)=a2<\/sup>V(X)<\/p>\n

          Question 27.
          \nFor a random variable V(X) = 4. Find V(3X +4) and V(-3X).
          \nAnswer:
          \nV(3X+4) = 32<\/sup> V(x) = 9 (4) = 36;
          \nV(-3X) = (-3)2<\/sup> V(x) = 9 (4) = 36; Since V(aX+b)=a2<\/sup> V(X),<\/p>\n

          Question 28.
          \nIf V(x)=1.6, then find V(2X – 5).
          \nAnswer:
          \nV(2X-5) = 22<\/sup> V(x) = 4 (1.6) = 6.4; Since V(aX+b)=a2<\/sup> V(X),<\/p>\n

          Question 29.
          \nIf E(X) = 5, E(Y) = 2, E(XY) = 11, then find Cov(X,Y).
          \nAnswer:
          \nWe know that Cov(X, Y) = E(XY) – E(X).E(Y)
          \n= 11 – (5) (2)=11 -10= 1.<\/p>\n

          Question 30.
          \nWhen X and Y are two discrete random variables with E(X)=5, E(Y)=6, E(X2<\/sup>)=34, E(Y2<\/sup>)=52, and E(XY)=25, then find Cov(X,Y) andin a bi-variate data
          \nAnswer:
          \nCov (X, Y) = E(XY)-E(X).E(Y) = 25 – (5) (6) = 25 – 30 = -5;
          \n\"1st<\/p>\n

          Question 31.
          \nIf V(X)=V(Y)=Cov(X,Y), then what is the value of ?
          \nAnswer:
          \n\u03b3 = 1<\/p>\n

          Question 32.
          \nIf P(X = 3) = 2\/3 and P(X = -2) = 1\/3, then find E(X) and V(3X + 4)
          \nAnswer:
          \nGiven probability distribution is:
          \n\"1st
          \nE(x) =8\/3;
          \nV(x) = E(X2<\/sup>)-[E(X)]2<\/sup>
          \n= – (8\/3)2<\/sup>
          \n= 22\/3-7.0756 = 7.3333-7.0756 = 0.2577
          \nV(3X + 4) = 32<\/sup>. V(X) = 9 (0.2577) = 2.3193<\/p>\n

          Question 33.
          \nIf P(X) is a probability mass function (p.m.f) then what is \u03a3P (x)?
          \nAnswer:
          \n\u03a3P(X)=1<\/p>\n

          Question 34.
          \nFind the mean and standard deviation for the following probability distribution.
          \n\"1st
          \nAnswer:
          \nWe Know that : Mean = E(x) : \u03a3x.P(X)=1
          \n\\(=(-2) \\times \\frac{1}{5}+(-1) \\times \\frac{2}{5}+0 \\times \\frac{1}{5}+2 \\times \\frac{1}{10}+4 \\times \\frac{1}{10}\\)
          \nE(x) =-0.4 -0.4 + 0 +0.2+ 0.4 = -0.2
          \nVar(x) = E(x2<\/sup>) – [E(x)]2<\/sup>
          \nHere E(x2<\/sup>) = [x2<\/sup>. P(x)]
          \n\\(=(-2)^{2} \\times \\frac{1}{5}+(-1)^{2} \\times \\frac{2}{5}+0^{2} \\times \\frac{1}{5}+2^{2} \\times \\frac{1}{10}+4^{2} \\times \\frac{1}{10}\\)
          \n= 0.8+ 0.4+ 0 + 0.4+ 1.5
          \nE(x2<\/sup>) = 3.2
          \n\u2234 Var(x) =3.2-(-0.2)2<\/sup> = 3.2 – 0.04 = 3.16
          \n\"1st<\/p>\n

          \"KSEEB<\/p>\n

          Question 35.
          \nIf two coins are tossed once. Find the mean and variance of number of Heads.
          \nAnswer:
          \nS = {HH, HT, TH, TT}
          \nThe probability of number of Heads is :
          \n\"1st
          \nE(x) = \u03a3x. P(x) = \\(0 \\times \\frac{1}{4}+1 \\times \\frac{2}{4}+2 \\times \\frac{1}{4}\\)
          \nVar (x) = E(X2<\/sup>) [E(x)]2<\/sup>
          \nE(x2<\/sup>) = Ex2<\/sup>. P(x) .
          \n\\(=0^{2} \\times \\frac{1}{4}+1^{2} \\times \\frac{2}{4}+2^{2} \\times \\frac{1}{4}=0+0.5+1=1.5\\)
          \n\u2234 Var (x) = 1.5 – 12<\/sup> = 0.5<\/p>\n

          Question 36.
          \nA fair die thrown once. A person will gets \u20b9 5 if the faces of a die results in mulitple of 2. other wise he loses \u20b9 2. find his expected amount.
          \nAnswer:
          \nLet x denote the expected amount is random variable which can take: 5 (die results in multiple of 2 : 2.46) and -2( die results in 1, 3, 5), with respective probabilities.
          \nP(x = 5) = P(getting multiple of 2) = \\(\\frac{3}{6}=\\frac{1}{2}\\)
          \nP(x = -2) = P(Not getting multiple 2) = \\(\\frac{3}{6}=\\frac{1}{2}\\)
          \nThe probability distribution is :
          \n\"1st
          \nExpected amount: E(x) = \u03a3x. P(x) = \\(5 \\times \\frac{1}{2}-2 \\times \\frac{1}{2}=2.5-1\\)<\/p>\n

          Question 37.
          \nA person tosses a fair coin, he gets \u20b9 50 if head turns up and loses \u20b9 10 if tail appears. Find the expected amount, also find the standard deviation.
          \nAnswer:
          \nLet x denote the person to gets the amount is a random variable, which takes the values 50 and -10, with respective probabilities:
          \nP(x = 50) = P(getting Head) = \\(\\frac { 1 }{ 2 }\\) and P(x = -10) = P(getting tail) = \\(\\frac { 1 }{ 2 }\\)
          \nThe probability distribution is :
          \n\"1st
          \nThe expected amount: E(x) = \u03a3x. P(x) = 50 \u00d7 \\(\\frac { 1 }{ 2 }\\) -10 \u00d7 \\(\\frac { 1 }{ 2 }\\) = 25 – 5 = \u20b920
          \n\"1st
          \nHere E(x2<\/sup>) = Ex2<\/sup>. P(x) = (50)2<\/sup> \u00d7 \\(\\frac { 1 }{ 2 }\\) + (-10)2<\/sup> \u00d7 \\(\\frac { 1 }{ 2 }\\)
          \nE(x2<\/sup>) = 1250 + 50= 1300
          \n\u2234 S.D(x) = \\(\\sqrt{1300-(20)^{2}}=\\sqrt{900}\\) =\u20b9 30.<\/p>\n

          Question 38.
          \nA box has 5 blue and 3 yellow marbles. Two marbles are drawn at random. Find the expected number of blue marbles drawn.
          \nAnswer:
          \nLet x denote the number of blue marbles drawn, is a random varible which takes the values: 0, 1,2. The respective probabilities are calcualted as below:
          \nP(x = 0) = P(drawing no blue marbles)
          \n= P(2 yellow marbles) = \\(\\frac{^{3} \\mathrm{C}_{2}}{^{8} \\mathrm{C}_{2}}=\\frac{3}{28}\\)
          \nP(x = 1) = P(drawing one blue and one yellow marbles)
          \n\\(=\\frac{^{5} \\mathrm{C}_{1} \\times^{3} \\mathrm{C}_{1}}{^{8} \\mathrm{C}_{2}}=\\frac{15}{28}\\)
          \nP(x = 2) = P(drawing two blue marbles) = \\(\\frac{^{5} \\mathrm{C}_{2}}{^{8} \\mathrm{C}_{2}}=\\frac{10}{28}\\)
          \nThen the probability distribution is :
          \n\"1st
          \nExpected number of blue marbles: E(x) = \u03a3x P(x)
          \n\\(=0 \\times \\frac{3}{28}+1 \\times \\frac{15}{28}+20 \\times \\frac{10}{28}=0+0.5357+0.3571\\)
          \n\u2234 E(x) =0.8928<\/p>\n

          Question 39.
          \nThe following is the probability distribution of x. Find k and E(x).
          \n\"1st
          \nAnswer:
          \nFor a probability mass funciton : EP(x) = 1
          \n\\(\\text { i.e., } \\frac{1}{4}+\\frac{1}{2}+\\frac{3}{10}+k+\\frac{1}{8}=1\\)
          \n\\(\\mathrm{k}=1-\\left[\\frac{1}{4}+\\frac{1}{2}+\\frac{3}{10}+\\mathrm{k}+\\frac{1}{8}\\right]\\) \u2234K = 1 – 0.8925 = 0.1075
          \n\u2234 The probability distribution is
          \n\"1st
          \nE(x) = \u03a3x. P(x) = (1 \u00d7 0.25) + (2 \u00d7 0.33) + (3 \u00d7 0.1875) + (4 \u00d7 0.1075) + (5 \u00d7 0.125) = 2.5275<\/p>\n

          Question 40.
          \nFrom the following probability distribution of x, Find k, E(x), E(x + 3) and Uar (x).
          \n\"1st
          \nAnswer:
          \nBy the definition of p.m.f: \u03a3 P(x) = 1
          \n\\(\\therefore \\quad \\frac{1}{5}+\\frac{2}{5}+\\frac{1}{5}+k=1\\)\u00a0 \\(\\left.\\mathrm{k}=-1-\\left[\\frac{1}{5}+\\frac{2}{5}+\\frac{1}{5}\\right]=0.2+0.4+0.2\\right]\\)
          \nk = 1 – 0.8 = 0.2
          \nTo find E(x), E(2x + 3), (x) and (2x + 3) we need compute from the probability distribution.
          \n\"1st
          \n\u2234 E(x) = \u03a3x. P(x) = 1.4
          \nVar (x) = E(x2<\/sup>) – [E(x)]2<\/sup> = \u03a3x2<\/sup>. P(x) – (1.4)2<\/sup>
          \n= 3 – 1.96= 1.04
          \nE(x + 3) = \u03a3(x + 3) P(x) = 7.4<\/p>\n

          Question 41.
          \nIf x is a discrete random variable and let a, b be two constants, show that
          \n1. E(a) = a
          \n2. E(ax) = aE(x)
          \n3. E(ax + b) = aE(x) + b
          \n4. Var (a) = 0
          \n5. Var (ax) = a2<\/sup> Var (x)
          \n6. Var (ax + b) = a2<\/sup> Var (x).
          \nAnswer:
          \nProof: From the definitions of probability mass function and mathematical expectation
          \n1. E(a) = \u03a3a .P(a) = a. \u03a3P(x) = a.l = a. \u2235\u03a3P(x) = 1<\/p>\n

          2. E(ax) = \u03a3ax P(x) = a\u03a3x P(x) = a. E(x) \u2235\u03a3x P(x) = E(x)<\/p>\n

          3. E(ax + b) = E(ax + b). P(x) Removing bracket
          \n= \u03a3ax. P(x) + \u03a3b.P(x) = a\u03a3x . P(x) + b\u03a3P(x)
          \n= a. E(x) + b. 1 = aE(x) + b<\/p>\n

          4. Var (a) = E[a – E(a)]2<\/sup>; \u2234 V(x) = E[x – E(x)]2<\/sup>
          \n= E[a – a]2<\/sup> = E(0) = 0<\/p>\n

          5. Var (an) = E[ax – E(ax)]2<\/sup> = E[ax – aE(x)]2<\/sup>; \u2234E(ax) = aE(x)
          \n= a2<\/sup>E(x – E(x)]2<\/sup> = a2<\/sup> Var (x)<\/p>\n

          6. Var (ax + b) = E[(ax + b) – E(ax + b)]2<\/sup>
          \n\"1st
          \n= E[ax – aE(x)]2<\/sup> = a2<\/sup>. E[x – E(x)]2<\/sup> = a2<\/sup> Var (x)<\/p>\n

          \"KSEEB<\/p>\n

          Question 42.
          \nDefine joint\/ Bivariate probability function.
          \nAnswer:
          \nLet x and y be two discrete random variables, then P(x, y) is called bivariate or joint probability function of x and y, if P[x = x] and P(y = y)<\/p>\n

          Question 43.
          \nDefine margined probability functionof x.
          \nAns.
          \nMargined probability function of x: P(x) is obtained from the joint probability faction of P(x,y).
          \n\"1st<\/p>\n

          Question 44.
          \nDefine marginal probability function of y.
          \nAns.
          \nMarginal probability function of y: P,(y) 1’s obtained from the joint probability function of P(x, y).
          \n\"1st<\/p>\n

          Question 45.
          \nState and prove the addition theorem of mathematical expectation. \u2019
          \nAnswer:
          \nStatement: Let x and y be random variables with respective expectations E(x) and E(y). Then, the expectations of the sum of the random variables is
          \nE(x + y) =E(x) + E(y)
          \nProof: Let x and y be two discrete random variables, with their joint probability function : P(x, y) as,
          \nP(x,y) =P(x = x, y = y)
          \nThe marginal probability distributions of x and y are:
          \n\"1st<\/p>\n

          Question 46.
          \nState tne prove multiplication theorem of mathematical expectation.
          \nAnswer:
          \nStatement: Let x and y be two independent discrete random variables with respective expectations E(x) and E(y). then the expectation of the product of these random variables is: E(xy) = E(x) . E(y)
          \nProof: Let x and y be two independent discrete random variables, then joint probability function P(x, y) is
          \nP(x, y) = P1<\/sub>(x). P2<\/sub>(y) ……(i)
          \nThen, by the defintion of mathematical expectation.
          \n\"1st
          \nFrom the definition mathematical expectation
          \nE(x y) = E(x) . E(y), Hence the proof.<\/p>\n

          Question 47.
          \nFind, Mean, Variance of the following probability distribution.
          \n\"1st
          \nAnswer:
          \nMean = E(x) = \u03a3x . P(x)
          \n= (-2) \u00d7 0.05 + (-1) 0.2 + 0 \u00d7 0.25 + 1 \u00d7 0.4 + 2 \u00d7 0.1
          \n= -0.1 – 0.2 + 0 + 0.4 + 0.2 = 0.3
          \nVar(x) =E(x2<\/sup>) – [E(x)]2<\/sup>
          \nE(x2<\/sup>) = \u03a3x2<\/sup>. P(x) = (-2)2<\/sup> x 0.05 + (-1)2<\/sup> x 0.2 + 0 x 0.25 + 1 x 0.4 + 2 x 0.1
          \n= 0.2 + 0.2 + 0 + 0.4 + 0.2 = 1
          \n\u2234 Var (x) = 1 – (0.3)2<\/sup> = – 0.09 = 0.91<\/p>\n

          Question 48.
          \nFor the following probability distribution find E(x), E(2x + 5), E(x2 + 3), V(x), V(2x).
          \n\"1st
          \nAnswer:
          \nE(x) = \u03a3x . P(x)
          \n= 1 \u00d7 0.3 + 2 \u00d7 0.4 + 3 \u00d7 0.3
          \n= 0.3 + 0.08 + 0.9 = 20
          \nand E(x2<\/sup>) = \u03a3x2<\/sup>. P(x)
          \n= 12<\/sup> \u00d7 0.3 + 22<\/sup> \u00d7 0.4 + 32<\/sup> \u00d7 0.3
          \n= 0.3 +1.6+ 2.7 = 4.6
          \nVar (x) = E(x2<\/sup>) – [E(x)]2<\/sup> = 4.6 – (2)2<\/sup> = 0.6
          \nE(2x + 5); From the results : E(ax + b) = aE(x) + b
          \n= 2E(x) + 5; Since E(x) = 2
          \nE(2x + 5)=2 \u00d7 2 + 5 = 9
          \nE(x2<\/sup> + 3) = E(x2<\/sup>) + 3 = 4.6 + 3 = 7.6
          \n\u2234 Var (2x) = 22<\/sup> Var (x) = 4 x 0.6 = 2.4<\/p>\n

          Question 49.
          \nFrom the Following probability distribution. find k, E(x), E\\(\\left(\\frac{x}{3}\\right)\\), E(-4x) Var (x) Var (5x) var \\(\\left(\\frac{2 x}{3}\\right)^{3}\\), Var (-3x).
          \n\"1st
          \nAnswer:
          \nWe know from the probability mass function. \u03a3P(x) = 1
          \n\\(\\therefore \\quad \\frac{\\mathrm{k}}{4}+\\frac{\\mathrm{k}}{8}+\\frac{\\mathrm{k}}{2}+\\frac{\\mathrm{k}}{8}=1\\) \\(k\\left[\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{2}+\\frac{1}{8}\\right]=1\\)
          \nK.1 = 1; \u2234K = 1
          \nThe probability distribution reduce now,
          \n\"1st
          \n\\(\\mathrm{E}(\\mathrm{x})=\\Sigma \\mathrm{x} . \\mathrm{P}(\\mathrm{x})=(-2) \\times \\frac{1}{4}+0 \\times \\frac{1}{8}+1 \\times \\frac{1}{2}+2 \\times \\frac{1}{8}\\)
          \n= -0.5 + 0+ 0.5 + 0.25
          \nE(x) = 0.25
          \n\\(\\mathrm{E}\\left(\\mathrm{x}^{2}\\right)=(-2)^{2} \\times \\frac{1}{4}+0^{2} \\times \\frac{1}{8}+1^{2} \\times \\frac{1}{2}+2^{2} \\times \\frac{1}{8}\\)
          \n= 1 + 0 + 0.5 + 0.5 = 2
          \nVar (x) = E(x2<\/sup>) – [E(x)]2<\/sup>
          \n= 2 – (0.25)2<\/sup> = 2 – 0.0625 = 1.9375
          \n\\(\\mathrm{E}\\left(\\frac{\\mathrm{x}}{3}\\right)=\\mathrm{E}\\left(\\frac{1}{3} \\cdot \\mathrm{x}\\right)\\); From E(ax) = aE(x)
          \n\\(=\\frac{1}{3} \\mathrm{E}(\\mathrm{x})=\\frac{1}{3} \\times 0.25=0.083\\)
          \nE(-4x) = -4E()x = -4 \u00d7 0.25 = -1
          \nVar (5x) = 52<\/sup> var (x); from Var (ax) = a2<\/sup> Var (x)
          \n= 25 \u00d7 1.9375 = 48.4375
          \nVar\\(\\left(\\frac{2 x}{3}\\right)\\) = Var\\(\\left(\\frac{2}{3} \\cdot x\\right)\\)
          \n= \\(\\left(\\frac{2}{3}\\right)^{2}\\)Var(x) = \\(\\frac { 4 }{ 9 }\\) x 1.9375 = 0.861
          \nVar (-3x) = (-3)2<\/sup> Var (x) = 9 \u00d7 1.9375 = 17.4375<\/p>\n

          \"KSEEB<\/p>\n

          Question 50.
          \nA purse has 5 Ten rupee coins, 3 five rupee coins and 2 two rupee coins. A coin is drawn at random from the purse. Find expectation of amount draw.
          \nAnswer:
          \nLet x denote the expected amount drawn, is a discrete random variable which takes the values : x = 10, 5, 2 with the prababilities
          \nP(x = 10) = P(drawing \u20b9 10 coin) = \\(\\frac { 5 }{ 10 }\\)
          \nP(x=5)= \\(\\frac { 3 }{ 10 }\\). P(x = 2)=\\(\\frac { 2 }{ 10 }\\)
          \nThen, the probability distribution is
          \n\"1st
          \nExpected amount draw : E(x) = Ex. P(x)
          \n\\(=10 \\times \\frac{5}{10}+5 \\times \\frac{3}{10}+2 \\times \\frac{2}{10}=\\frac{50}{10}+\\frac{15}{10}+\\frac{4}{10}\\) = \u20b9 6.90<\/p>\n

          Question 51.
          \nIn a lottery there are 1000 tickets costing \u20b95 each. There is one first prize worth \u20b9 1000. Two second prizes worth \u20b9 500 each and ten third prizes worth \u20b9 100 each. Find expected gain in buying one lottery ticket.
          \nAnswer:
          \nLet x denote the expected gain is a discrete random variable which takes the values x : 995, 495, 95 and -5. Since he buys a ticket of for \u20b9 5 is deducted). The respective probability can be calculated as below:
          \np(x = 995) = \\(\\frac { 1 }{ 1000 }\\): p(x = 495) = \\(\\frac { 2 }{ 1000 }\\)
          \np(x = 95) = \\(\\frac { 10 }{ 1000 }\\) p(x = -5) = \\(\\frac { 987 }{ 1000 }\\)
          \nThe probability distribution is :
          \n\"1st
          \nExpected gain : E(x) = \u03a3x. P(x)
          \n\\(=995 \\times \\frac{1}{1000}+495 \\times \\frac{2}{1000}+95 \\times \\frac{10}{1000}-5 \\times \\frac{987}{1000}\\)
          \n= 0.995 + 0.99 + 0.95 – 4.985 = -2.47
          \ni.e., expected loss in buying a lottery ticket is \u20b9 247.<\/p>\n

          Question 52.
          \nA person enters into a fair by paying \u20b910. In the fair if choose to hit a target with airgun by paying \u20b9 1 per shot if he hits gets \u20b9 50. The probability of hitting a target is 1\/2. Find the expect amount.
          \nAnswer:
          \nLet x denote the expect amount is a discrete random variable, which takes the values : -11, 39.
          \nThe respective probabilities
          \nP(x = -11) = P(not hitting a target) = \\(\\frac { 2 }{ 3 }\\)
          \nP(x = 39) = P(hitting a target) = \\(\\frac { 1 }{ 3 }\\)
          \nTherefore, the expected amount is :
          \nE(x) = \u03a3x. P(x)
          \n= -11 \u00d7 \\(\\frac { 2 }{ 3 }\\) + 39 \u00d7 \\(\\frac { 1 }{ 3 }\\) = -7.33 + 13 = Rs. 5.67 profit.<\/p>\n

          Question 53.
          \nFind expected sum of numbers obtained in a throw of two dice.
          \nAnswer:
          \nLet x and y be the numbers appeared on the faces of two dice is a discrete random variable, then each die show with numbers 1, 2,3, 4, 5, 6 with respective probabilities \\(\\frac { 1 }{ 6 }\\) each.
          \nExpected number obtained on the first die be : E( x) = \u03a3x . P(x)
          \n\\(=1 \\times \\frac{1}{6}+2 \\times \\frac{1}{6}+3 \\times \\frac{1}{6}+\\ldots \\ldots+6 \\times \\frac{1}{6}=3.5\\)
          \nSimilarly, expected number on the faces of 2nd<\/sup> die be E(y) = 3.5
          \nThe expectation of sum of numbers obtained is
          \nE(x + y) = E(x) + E(y) = 3.5 + 3.5 = 7
          \nNote: Similarly, sum of’ ‘n’ dice of numbers can be calcualted.<\/p>\n

          Question 54.
          \nA bag has ten tiny of dice. Two dice are randomly drawn. Find the expectation of the product of the numbers obtained on the faces of dice.
          \nAnswer:
          \nLet x and y denoted the numbers obtained on the faces of dice, is a discrete random variable with values 1, 2,…. 6 and with the probability \\(\\frac { 1 }{ 6 }\\) each.
          \nTherefore E(x) = \\(1 \\times \\frac{1}{6}+2 \\times \\frac{1}{6}+\\ldots \\ldots+6 \\times \\frac{1}{6}=3.5\\)
          \nSimilarly, E(y) =3.5
          \nSince, drawing dice are independens, therefore
          \nE(x, y) = E(x) . E(y) = 3.5 \u00d7 3.5 = 12.25
          \nNote: Similarly product of’n’ dice of numbers can also be calculated in the square way.<\/p>\n

          Question 55.
          \nIf in a bivoriate data, E(x) = 4, E(x2<\/sup>) = 25, E(y) = 8 and E(y2<\/sup>) = 100 and E(x y ) = 24. Find Var (x), Var (y), Cov (x, y) and the coefficient of correlation (3xy).
          \nAnswer:
          \nVar (x) = E(x2<\/sup>) – [E(x)]2<\/sup> = 25 – 42<\/sup> = 9
          \nVar (x) = E(y2<\/sup>) – [E(y)]2<\/sup> = 100 – 82<\/sup> = 36
          \nCOV (x, y) = E(xy) – E(x) . E(y) = 24 – 4 x 8 = -8
          \n\"1st<\/p>\n

          Question 56.
          \nIn a bivariate data E(x) = 1.3. E(x2<\/sup>) = 3.4, E(y) = 0 and E(y2<\/sup>) = 4. E(xy) = 0.1 Find S.D(x), SD(y) and rxy.
          \nAnswer:
          \n\"1st<\/p>\n

          Question 57.
          \nFor the following bivariate probaiblity distribution, find E(x + y) and E(xy).
          \n\"1st
          \nAnswer:
          \n\"1st
          \nHere 0.08, 0.17, 0.25, 0.08, 0.25 and 0.17 are joint probability function P(x, y) of discrete random variables x and y.
          \nE(x.y) =(-2) \u00d7 0 \u00d7 0.08 + 0 \u00d7 0 \u00d7 0+1 \u00d7 0 \u00d7 0.25 + (-2) \u00d7 1 \u00d7 0.17 + 0 \u00d7 1 \u00d7 0.25 + 1 \u00d7 1 \u00d7 0.17 +(2) \u00d7 2 \u00d7 0 + 0 \u00d7 2 \u00d7 0.08+ 1 \u00d7 2 \u00d7 0.
          \n= (0 + 0 + 0) +(0.34 + 0 + 0.17) + (0 + 0 + 0)
          \n\u2234 E(x.y) =0.17<\/p>\n

          \"1st
          \n= (-2 + 0) 0.08 + (0 + 0) 0 + (1 + 0) 0.25 + (-2 + 1) 0.17 + (0 + 1)0.25 + (1 + 1)0.17+ (-2 +2) +0 + (0 + 2) 0.08+ (1 +2)0 = (-0.16 + 0 + 0.25) + (-0.17 + 0.25 + 0.34) + (0 + 0.16 + 0)
          \n= 0.09 + 0.42 + 0.16.
          \nE(x + y) =0.67<\/p>\n

          \"KSEEB<\/p>\n

          Question 58.
          \nFrom the following bivariate probability distribution, find the coefficient of correlation.
          \n\"1st
          \nAnswer:
          \nThe coefficient of correlation is :
          \n\"1st
          \nHere Var(x) = E(x2<\/sup>) – [E(x)]2<\/sup>
          \n‘To calculate, obtain the marginal probility distribution of x:
          \n\"1st
          \nHere : 0.05 + 0.25 + 0.1 = 0.4 and 2 : 0.4 + 0 + 0.2 = 0.6
          \nE(x) = Ex . P(x) = 1 \u00d7 0.4 + 2 \u00d7 0.6 = 1.6
          \nE(x2<\/sup>) = Ex2<\/sup>. P(x) = 12<\/sup> \u00d7 0.4 + 22<\/sup> \u00d7 0.06 = 2.8
          \n\u2234 Var (x)= 2.8 \u2014 (1.6)2<\/sup> = 0.24<\/p>\n

          The marginal probability distribution of y:
          \n\"1st
          \nE(y) = Ey P(y) = (-1) \u00d7 0.45 + (-2) \u00d7 0.25 + 0 \u00d7 0.3 = – 0.95
          \nE(y2) = (-1 )2<\/sup> \u00d7 0.45 + (-2)2<\/sup> \u00d7 0.25 + 02<\/sup> \u00d7 0.3 = 0.45 + 1 + 0 = 1.45
          \n\u2234 Var(y) = E(y2<\/sup>) – [E(y)]2<\/sup> = 1.45 – (-0.95)2<\/sup> = 1.45 – 0.9025 = 0.5475
          \ncov (x, y) = E(xy) – E(x) . E(y)
          \n\"1st
          \n= (1 \u00d7 (-1) \u00d7 0.05) + (2 \u00d7 (-1) \u00d7 0.4) + (1 \u00d7 (-2) \u00d7 0.25) + (2 \u00d7 (-2) \u00d7 0) + (1 \u00d7 0 \u00d7 0.1) +(2 \u00d7 0 + 0.2)
          \n= -0.05 -0.8 -0.5 + 0 + 0 + 0
          \nE(xy) =-1.35
          \n\u2234 cov(x,y) =-1.35- 1.6 x (-0.95) = -1.35 + 1.52 = -0.17
          \n\"1st<\/p>\n

          Question 59.
          \nIf E(x2<\/sup>) = 65, and E(x) = 4. Find standard deviation of x.
          \nAnswer:
          \nHere,
          \n\"1st
          \nVar(x) = E(x2<\/sup>) = [E(x)]2<\/sup> = 65 – (42<\/sup>) = 65 – 16 = 49
          \n\"1st<\/p>\n

          Question 60.
          \nIfVar(x)=0.4327, and E(x2) = 0.75. Find E(x).
          \nAnswer:
          \nVar(x) = E(x2<\/sup>) – [E(x)]2<\/sup>
          \n0.4327 = 0.75 – [E(x)]2<\/sup>
          \n[E(x)]2<\/sup> =0.75-0.4327 = 0.3153
          \n\u2234 E(x) = \\(\\sqrt{0.3153}\\) =0.5615<\/p>\n

          Question 61.
          \nIf x and y be two independent random variable with E(xy) = 8, and E(x) = 2, then find E(y).
          \nAnswer:
          \nSince x and y are independent then E(xy) = E(x). E(y)
          \ni.e., 8 = 2. E(y) \u2234 E(y) = 2<\/p>\n

          Question 62.
          \nFor a bivariate data if E(x) = 5, E(y) = 3, E(x2<\/sup>) = 45, E(y2<\/sup>) = 85 and E(x) = 18. Find yxy.
          \nAnswer:
          \n\"1st
          \nso, Var (x) = E(x2<\/sup>) = [E(x)]2<\/sup> = 45 – (5)2<\/sup> = 45 – 25 = 20
          \nVar (y) = E(y2<\/sup>) – [(y)]2<\/sup> = 85 – (3)2<\/sup> = 85 – 9 = 76
          \nCOV (x, y) = E(xy) – E(x) . E(y) =18-5 x 3 = 18- 15 = 3
          \n\u2234 \\(\\gamma=\\frac{3}{\\sqrt{20} \\times \\sqrt{76}}=\\frac{3}{4.472 \\times 8.718}=\\frac{3}{38.987}=0.0769\\)<\/p>\n

          Question 63.
          \nIf E(x) = 5 and E(y) = 12, find E(x + y), E(x – y), E(4x + 2y), and E(-3x + 2y).
          \nAnswer:
          \nE(x + y) = E(x) + E(y) , by theorem = 5 + 12= 17
          \nE(x-y) = E(x) – E(y) = 5 – 12 = -7
          \nE(4x + 2y) = 4E(x) + 2E(y) = 4 \u00d7 5 + 2 \u00d7 12 = 20 + 24 = 44
          \nE(-3x + 2y) = (-3) E(x) + 2E(y) = (-3) \u00d7 5 + 2 \u00d7 12 = 15 + 24 = 9<\/p>\n

          Question 64.
          \nIf Var (x) = 3, find Var (5x), Var (-8x), Var (-x), Var (4x – 5) and Var (8 – 3x)
          \nAnswer:
          \nV(5x); By the results ; Var (ax) = a2<\/sup> Var (x)
          \n\u2234 Var (5x) = 52<\/sup> Var (x) = 25 x 3 = 75
          \nVar (-8x) = (-8)2<\/sup> Var (x) = 64 x 3 = 192
          \nVar (-x) = Var (-1 .x) = (-1)2 Var (x) = 1 x 3 = 3.
          \nVar (4x – 5); using the result; Var (ax + b) = a2<\/sup> Var (x)
          \n\u2234 Var(4x – 5) = 42<\/sup> Var (x) = 16 x 3 = 48
          \nVar (8 – 3x) = Var (-3x + 8) = (-3)2<\/sup> Var (x) = 9 x 3= 27<\/p>\n

          \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

          You can Download Chapter 10 Random Variables and Mathematical Expectation Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Statistics Question Bank Chapter 10 Random Variables and Mathematical Expectation 1st PUC Random Variables …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12033"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=12033"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12033\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=12033"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=12033"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=12033"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}