Probability of getting head and tail are denoted by p

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Probability of getting head and tail are denoted by p and q respectively. In case of unbiased coin p = q = 1/2. The probability of r success i.e. getting heads in 8 trails is given by P(r)= nc r q n r p r r r 8 r r r - 8 r 2 1 2 1 8c p q 8c r 8 r 8c 256 1 2 1 8c Probability of obtaining at least six heads, 1] 8 [28 256 1 ] 8c 8c [8c 256 1 8 7 6 256 37 Hence the probability of obtaining at least six heads is 37/256.
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37 Illustration - 13 The incidence of occupational disease in an industry is such that the workman have a 20% chance of suffering from it. What is the probability that out of six workman, three or more will contact disease. Solution Probability of a man suffering from disease is 5 1 100 20 p Therefore, q = 1 p = 1 1/5 = 1/4 Probability of 3 or more men suffering from disease is termed in binomial expansion of 6 5 1 5 4 Hence the probability of 3 or more success i.e. 3, 4, 5 or 6 success is : = nc 3 q 3 p 3 + nc 2 q 4 p 2 + nc 1 q 5 p + p 6 = 20 q 3 p 3 + 15p 4 p 2 + 6q 5 p + p 6 6 5 4 2 3 3 5 1 5 1 5 4 6 5 1 5 4 15 5 1 5 4 20 6 5 4 2 3 3 5 1 3 4 6 3 1 3 4 15 5 1 5 4 20 15625 1 15625 4 6 15625 16 15 15625 64 20 09888 . 0 15625 1545 Illustration - 14 a) For a binomial distribution, mean = 7 and variance = 11. Comment b) If the probability of a defective item is 1/12, Find (i) the mean, (ii) variance. Solution a) We are given, Mean = 7 and Variance = 11. We know that Mean = np, variance = npq We get the value of p by substituting the given values in 57 . 1 7 11 np npq
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38 b) We are given, n = 600, p = 1/12 and q = 1 1/12 = 11/12 By using binomial law: Mean = np = 600 12 1 = 50 Variance = npq = 600 12 11 12 1 = 45.83 Illustration - 15 Nine coins are tossed at a time, 512 times. Number of head observed at each throw is recorded and the results are given below. Find the expected frequencies. What are the theoretical values of mean and standard deviation. Also calculate the mean and standard deviation of the observed frequencies. No. of heads at a throw Frequency 0 1 2 3 4 5 6 7 8 9 4 10 45 115 139 105 65 19 8 2 Solution Probability of getting a head in a single throw = 1/2. Therefore, q = 1 p = 1 1/2 = 1/2 Now we have, p = 1/2, q = 1/2, n = 9, N = 512 The expected frequencies can be calculated by expanding 512 (1/2 + 1/2) 9
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