SOLUTION: Suppose the scores of students on an exam are Normally distributed with a mean of 488 and a standard deviation of 67. Then approximately 99.7% of the exam scores lie between the nu
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Question 1200611: Suppose the scores of students on an exam are Normally distributed with a mean of 488 and a standard deviation of 67. Then approximately 99.7% of the exam scores lie between the numbers
and
such that the mean is halfway between these two integers. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! mean is 488 and standard deviation is 67.
99.7% of the exam scores lie between 289.162 and 686.838
critical z-score for two tailed confidence interval os .997 is equal to plus or minus 1.967737927.
low z-score formula becomes -2.967737927 = (x - 488) / 67.
solve for x to get x = 67 * -2.967737927 + 488 = 289.163
high z-score formula becomes 2.967737927 = (x - 488) / 67.
solve for x to get x = 67 * 2.967737927 + 488 = 686.838.
i did it the easy way by using the calculator at https://davidmlane.com/hyperstat/z_table.html
here are the results from using that calculator.
