C question 4 project management and random variables

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c) Question 4 – Project Management and Random Variables
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a) Mean = 145 Variance = 36 Standard Deviation = 6 b) i) Pr ( Z > 3.33) = 0.9996 1-0.9996 = 0.0004 or 0.04% ii) Pr ( Z > -0.67) = 0.2514 or 25% iii) Pr ( -0.67 < Z < 3.33) = 0.7482 or 75% iv) Pr ( Z > ? ) = 0.05 1-0.05 = 0.9500 == z = 1.645 X = 145 + 1.645*6 = 154.87 or 155 minutes Question 5 – Sampling Distribution of Proportions a) The general shape of the sampling distribution of the sample proportion of customers who are satisfied is approximately normal. We can assume that because X is a binomial distribution and we can use a normal distribution to approximate it. Mean () = p b) Pr ( Z < -2.45) = 0.0071 or 0.71% c) Pr ( Z < -3.46) = 0.0003 or 0.03% d) The bookstore’s claim is most likely false because the probability of having less than 45% satisfied customers was extremely low, 0.71%, which means that the bookstore’s true proportion of 50% is wrong. Question 6 – Sampling Distribution of Means
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a) Mean = 604.02 St. Dev = 64.05 b) The general shape of the sampling distribution of the sample mean is approximately normal. The mean of the sample mean is 604.02. The standard deviation of the sample mean is 9.77. c) Pr ( Z < -0.41) = 0.3409 or 34% d) Mean = 633.23 St. Dev = 103.29 The general shape of the sampling distribution of the sample mean is approximately normal. The mean of the sample mean is 633.23. The standard deviation of the sample mean is 9.99. e) Pr ( Z > -2.09) = 0.0183 1-0.0183 = 0.9817 or 98%
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