c)
Question 4 – Project Management and Random Variables
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Mean = 145
Variance = 36
Standard Deviation = 6
b)
i) Pr ( Z > 3.33) = 0.9996
10.9996 = 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
10.05 = 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
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
10.0183 = 0.9817 or 98%
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 Winter '10
 E.Fowler
 Correlation, Normal Distribution, Standard Deviation, RIM, linear relationship

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