Exam2_practice1_solutions

Exam2_practice1_solutions - 5E practice problems for exam 2...

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Unformatted text preview: 5E practice problems for exam 2 1. The normal approximation will have the same mean and standard deviation as the binomial. These are 600 · (1 / 6) = 100 and p 600(1 / 6)(1- 1 / 6) = 9 . 13. We want the area to left of 80, so we calculate the z-score of 80: z = 80- 100 9 . 13 =- 2 . 19. The table gives .0143. A normal distribution is symmetric about its mean, so the area to the right of 120 is the same as the area to the left of 80: .0143. To find the probability that at least 110 are ones, we find the z-score of 110: z = 110- 100 9 . 13 = 1 . 10. From symmetry, we can use the table value for z =- 1 . 10 directly (and must use it for the half-table given in the exam). This value is .1357. 2. We could calculate each z-score in one fell swoop, but it will be convenient to first calculate the mean and standard deviation of the distribution of sample means. They are 22 and 8 / √ 5 = 3 . 58....
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