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# qt - Alternative Example 2.9 Here is how the expected...

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Unformatted text preview: Alternative Example 2.9: Here is how the expected outcome can be computed for the question in Alternative Example 2.8. _ x 3 P ( x 3 ) _ x 4 P ( x 4 ) _ x 5 P ( x 5 ) _ 5(0.4) _ 4(0.3) _ 3(0.2) _ 2(0.1) _ 1(0) _ 4.0 Alternative Example 2.10: Here is how variance is computed for the question in Alternative Example 2.8: _ (5 _ 4) 2 (0.4) _ (4 _ 4) 2 (0.3) _ (3 _ 4) 2 (0.2) _ (2 _ 4) 2 (0.1) _ (1 _ 4) 2 (0) _ (1) 2 (0.4) _ (0) 2 (0.3) _ ( _ 1) 2 (0.2) _ ( _ 2) 2 (0.1) _ 0.4 _ 0.0 _ 0.2 _ 0.4 _ 0.0 _ 1.0 The standard deviation is _ 1 Alternative Example 2.11: The length of the rods coming out of our new cutting machine can be said to approximate a normal distribution with a mean of 10 inches and a standard deviation of 0.2 inch. Find the probability that a rod selected randomly will have a length a. of less than 10.0 inches b. between 10.0 and 10.4 inches c. between 10.0 and 10.1 inches d. between 10.1 and 10.4 inches e. between 9.9 and 9.6 inches f. between 9.9 and 10.4 inches g. between 9.886 and 10.406 inches First compute the standard normal distribution, the Z-value: Next, find the area under the curve for the given Z-value by using a standard normal distribution table. a. P ( x _ 10.0) _ 0.50000 b. P (10.0 _ x _ 10.4) _ 0.97725 _ 0.50000 _ 0.47725 c. P (10.0 _ x _ 10.1) _ 0.69146 _ 0.50000 _ 0.19146 d. P (10.1 _ x _ 10.4) _ 0.97725 _ 0.69146 _ 0.28579 e. P (9.6 _ x _ 9.9) _ 0.97725 _ 0.69146 _ 0.28579 f. P (9.9 _ x _ 10.4) _ 0.19146 _ 0.47725 _ 0.66871 g. P (9.886 _ x _ 10.406) _ 0.47882 _ 0.21566 _ 0.69448 S OLUTIONS TO D ISCUSSION Q UESTIONS AND P ROBLEMS 2-1. There are two basic laws of probability. First, the probability of any event or state of nature occurring must be greater than or equal to zero and less than or equal to 1. Second, the sum of the simple probabilities for all possible outcomes of the activity must equal 1. z x _ _ = 1 = variance variance _ _ i x i E x P x i =- 1 5 ( ( )) 2 ( ) E x x P x x P x x P x i i i ( ) = ( ) = ( ) + ( ) = _ 1 5 1 1 2 2 2-2. Events are mutually exclusive if only one of the events can occur on any one trial. Events are collectively exhaustive if the list of outcomes includes every possible outcome. An example of mutually exclusive events can be seen in flipping a coin. The outcome of any one trial can either be a head or a tail. Thus, the events of getting a head and a tail are mutually exclusive because only one of these events can occur on any one trial. This assumes, of course, that the coin does not land on its edge. The outcome of rolling the die is an example of events that are collectively exhaustive. In rolling a standard die, the outcome can be either 1, 2, 3, 4, 5, or 6. These six outcomes are collectively exhaustive because they include all possible outcomes. Again, it is assumed that the die will not land and stay on one of its edges....
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qt - Alternative Example 2.9 Here is how the expected...

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