Asked by roger
Suppose the average body temp is 98.6 assume the variable is normally distributed with a standard deviation of .74 degrees. suppose a sample of 100 is taken. what is the probability that the mean is below 98.4??? Between 98.5 AND 98.7??
Answers
Answered by
Kuai
z = (98.5-98.6)/.74/sqr(100)
z = (98.7- 98.6)/.74/sqr(100)
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