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Unformatted text preview: &¡¢£¢ ¤¥ &¦§¨©ª¨«¬­®«¯­¦°± ¡­²³©ª­´µ 4.1. The random variable X is not a Binomial random variable, since the balls are selected without replacement (after we pick a ball, we do not put it back). For this reason, the probability p of choosing a red ball changes from trial to trial. 4.2. If the sampling in Exercise 4.1 is conducted with replacement (we pick a ball, note the color and put it back), then X is a Binomial random variable with n = 2 independent trials, and p = P [red ball] = 3/5, which remains constant from trial to trial. 4.5. With n = 6 and three different values of p , the values of P (X = k) = k k k q p C-6 6 are calculated for k = 0, 1, 2, 3, 4, 5, and 6. The values of P (X = k) are shown in the table below. p =.1 p =.5 p =.9 k P (X = k) P (X = k) P (X = k) 0 0.531441 0.03125 0.00032 1 0.352494 0.15625 0.0064 2 0.098415 0.3125 0.0512 3 0.01458 0.3125 0.2048 4 0.001215 0.15625 0.4096 5 0.000054 0.03125 0.32768 6 0.000001 The probability histogram is given below. Notice that the probability distribution for p = .1 and p = .9 are mirror images. When p is small, the distribution is skewed to the left, when p is large, the distribution is skewed to the right. 4.18. The random variable X is approximately Binomial, with n = 25 and probability of success p = P [CEO is aware of information superhighway] = .5. Using Table 1 from the textbook, for n= 25, we get: a. P [X = 25] = P [X ≤ 25] – P [X < 25] = P [X ≤ 25] - P [X ≤ 24] = 1 – 1 = 0 b. P [X ≥ 10] = 1 – P [X ≤ 9] = 1 – .115 = .885 c. P [ X = 10] = P [ X ≤ 10 ] – P [X < 10] = P [ X ≤ 10 ] - P [X ≤ 9 ] = .212 - .115 = .097 Note: we can calculate the probabilities using the formula for Binomial probability; however, for parts (b) the calculation will be very long (table is more convenient in this case). See below part (c), using the formula: [ ] 15 10 10 25 10 ) 5 . ( ) 5 . ( ! 15 ! 10 ! 25 ) 5 . 1 ( 5 . 10 25 10 =-& & ¡ ¢ £ £ ¤ ¥ = =-X P = 0.09 4.20. a. Let X = number of Japanese who feel their products are superior to American products. Then, X has a Binomial distribution (the alternative outcomes are ‘feel products are superior’ and ‘don’t feel products are superior’), with n = 50 and probability of success p = P [Japanese feel their products are superior] = .71. That is, P (X = k) = k k k C-50 50 ) 29 (. ) 71 (. for k = 0,1,2,…,50 b. If X = number of Japanese who feel the US would be the number one economic power in the next century, then X has a Binomial distribution with n = 50 and p = P [Japanese feel US will be #1 economic power] = .42. That is, P (X = k) = k k k C-50 50 ) 58 (. ) 42 (. for k = 0,1,2,…,50 c. For the random variable X described in part b , μ = np = 50(.42) = 21 and σ = npq = ) 58 )(. 42 (. 50 ( = 3.490 4.21....
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