{"id":35497,"date":"2022-09-19T17:01:15","date_gmt":"2022-09-19T11:31:15","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35497"},"modified":"2022-09-19T17:01:15","modified_gmt":"2022-09-19T11:31:15","slug":"inter-2nd-year-maths-2a-random-variables-and-probability-distributions-solutions-ex-10a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-random-variables-and-probability-distributions-solutions-ex-10a\/","title":{"rendered":"Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions<\/a> Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nA p.d.f of a discrete random variable is zero except at the points x = 0, 1, 2. At these points it has the value P(0) = 3c3<\/sup>, P(1) = 4c – 10c2<\/sup>, P(2) = 5c – 1 for some c > 0. Find the value of c.
\nSolution:
\nP(X = 0) + P(X = 1) + P(X = 2) = 1
\n3c3<\/sup> + 4c – 10c2<\/sup> + 5c – 1 = 1
\n3c3<\/sup> – 10c2<\/sup> + 9c – 2 = 0
\nPut c = 1, then 3 – 10 + 9 – 2 = 12 – 12 = 0
\nc = 1 satisfies the above equation
\nc = 1 \u21d2 P(X = 0) = 3 which is not possible
\nDividing with c – 1, we get
\n3c2<\/sup> – 7c + 2 = 0
\n(c – 2) (3c – 1) = 0
\n\u2234 c = 2 or c = \\(\\frac{1}{3}\\)
\nc = 2
\n\u21d2 P(X = 0) = 3 . 23<\/sup> = 24 which is not possible
\n\u2234 c = \\(\\frac{1}{3}\\)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the constant C, so that F(x) = \\(C\\left(\\frac{2}{3}\\right)^x\\), x = 1, 2, 3,……… is the p.d.f of a discrete random variable X.
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\n\"Inter
\nis the probability distribution of a random variable X. Find the value of K and the variance of X.
\nSolution:
\nSum of the probabilities = 1
\n\u21d2 0.1 + k + 0.2 + 2k + 0.3 + k = 1
\n\u21d2 4k + 0.6 = 1
\n\u21d2 4k = 1 – 0.6 = 0.4
\n\u21d2 k = \\(\\frac{0.4}{4}\\) = 0.1
\nMean = (-2) (0.1) + (-1) (k) + 0 (0.2) + 1 (2k) + 2(0.3) + 3k
\n= – 0.2 – k + 0 + 2k + 0.6 + 3k
\n= 4k + 0.4
\n= 4(0.1) + 0.4
\n= 0.4 + 0.4
\n= 0.8
\n\u03bc = 0.8
\nVariance (\u03c32<\/sup>) = \\(\\sum_{i=1}^n x_i^2 P\\left(x=x_i\\right)-\\mu^2\\)
\n\u2234 Variance = 4(0.1) + 1(k) + 0(0.2) + 1 (2k) + 4 (0.3) + 9k – \u03bc2<\/sup>
\n= 0.4 + k + 0 + 2k + 4(0.3) + 9k – \u03bc2<\/sup>
\n= 12k + 0.4 + 1.2 – (0.8)2<\/sup>
\n= 12(0.1) + 1.6 – 0.64
\n= 1.2 + 1.6 – 0.64
\n= 2.8 – 0.64
\n= 2.16
\n\u2234 \u03c32<\/sup> = 2.16<\/p>\n

Question 4.
\n\"Inter
\nis the probability distribution of a random variable X. Find the variance of X.
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 5.
\nA random variable X has the following probability distribution.
\n\"Inter
\nFind (i) k (ii) the mean and (iii) P(0 < X < 5)
\nSolution:
\nSum of the probabilities = 1
\n\u21d2 0 + k + 2k + 2k + 3k + k2<\/sup> + 2k2<\/sup> + 7k2<\/sup> + k = 1
\n\u21d2 10k2<\/sup> + 9k = 1
\n\u21d2 10k2<\/sup> + 9k – 1 = 0
\n\u21d2 10k2<\/sup> + 10k – k – 1 = 0
\n\u21d2 10k(k + 1) – 1(k + 1) = 0
\n\u21d2 (10k – 1) (k + 1) = 0
\n\u21d2 k = \\(\\frac{1}{10}\\), -1 Since k > 0
\n\u2234 k = \\(\\frac{1}{10}\\)<\/p>\n

(i) k = \\(\\frac{1}{10}\\)<\/p>\n

(ii) Mean = \\(\\sum_{i=1}^n x_i P\\left(x=x_i\\right)\\)
\nMean (\u03bc) = 0(0) + 1(k) + 2(2k) + 3(2k) + 4(3k) + 5(k2) + 6(2k2<\/sup>) + 7 (7k2<\/sup> + k)
\n= 0 + k + 4k + 6k + 12k + 5k2<\/sup> + 12k2<\/sup> + 49k2<\/sup> + 7k
\n= 66k2<\/sup> + 30k
\n= 66(\\(\\frac{1}{100}\\)) + 30(\\(\\frac{1}{10}\\))
\n= 0.66 + 3
\n= 3.66<\/p>\n

(iii) P(0 < x < 5)
\nP(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\n= k + 2k + 2k + 3k
\n= 8k
\n= 8(\\(\\frac{1}{10}\\))
\n= \\(\\frac{4}{5}\\)<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nThe range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c3<\/sup>, P(X = 1) = 4c – 10c2<\/sup>, P(X = 2) = 5c – 1
\n(i) Find the value of c
\n(ii) P(X < 1), P(1 \u2264 X < 2) and P(0 < X \u2264 3)
\nSolution:
\nP(X = 0) + P(X = 1) + P(X = 2) = 1
\n3c3<\/sup> + 4c – 10c2<\/sup> + 5c – 1 = 1
\n3c3<\/sup> – 10c2<\/sup> + 9c – 2 = 0
\nc = 1 satisfy this equation
\nc = 1 \u21d2 P(X = 0) = 3 which is not possible
\nDividing with c – 1, we get
\n3c2<\/sup> – 7c + 2 = 0
\n(c – 2) (3c – 1) = 0
\nc = 2 or c = \\(\\frac{1}{3}\\)
\nc = 2 \u21d2 P(X = 0) = 3 . 23<\/sup> = 24 which is not possible
\n\u2234 c = \\(\\frac{1}{3}\\)<\/p>\n

(i) P(X < 1) = P(X = 0)
\n= 3 . c3<\/sup>
\n= 3 . \\(\\left(\\frac{1}{3}\\right)^3\\)
\n= 3 . \\(\\frac{1}{2}\\)
\n= \\(\\frac{1}{9}\\)<\/p>\n

(ii) P(1 < X \u2264 2) = P(X = 2)
\n= 5c – 1
\n= \\(\\frac{5}{3}\\) – 1
\n= \\(\\frac{2}{3}\\)<\/p>\n

(iii) P(0 < X \u2264 3) = P(X = 1) + P(X = 2)
\n= 4c – 10c2<\/sup> + 5c – 1
\n= 9c – 10c2<\/sup> – 1
\n= 9 . \\(\\frac{1}{3}\\) – 10 . \\(\\frac{1}{9}\\) – 1
\n= 3 – \\(\\frac{10}{9}\\) – 1
\n= \\(\\frac{8}{9}\\)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nThe range of a random variable X is {1, 2, 3, …..} and P(X = k) = \\(\\frac{C^K}{K !}\\), (k = 1, 2, 3, ……), Find the value of C and P(0 < X < 3)
\nSolution:
\nSum of the probabilities = 1
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a) I. Question 1. A p.d.f of a discrete random variable is zero …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35497"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35497"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35497\/revisions"}],"predecessor-version":[{"id":35498,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35497\/revisions\/35498"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35497"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35497"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}