mba 522 Normal Example 3

# mba 522 Normal Example 3 - these values are found using Z...

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Let’s assume the average cost for a family of four to attend a Washington Redskins football game is \$442.54. Also assume the normal distribution applies and that the standard deviation is \$65. What is the probability that a family of four will spend between \$400 and \$500? X 400 442.54 500 Z -0.65 0 0.88
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Unformatted text preview: these values are found using Z formula. Dr. J. The shaded area is what would answer the question. It finds the area between \$400 and \$500. To compute the answer, do the following: 0.8106 – 0.2578 = 0.5528 This tail is 0.2578 Table says this whole area to the left of Z = 0.88 equals 0.8106....
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## This note was uploaded on 11/13/2011 for the course MBA 522 taught by Professor Nabavi during the Spring '08 term at Bellevue.

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