SOLUTION: assume body temp of healthy adults are normally distributed with a mean of 98.20F and a standard deviation of 0.62F. what body temperature is the 95th and 5th percentile??

Question 453135: assume body temp of healthy adults are normally distributed with a mean of 98.20F and a standard deviation of 0.62F. what body temperature is the 95th and 5th percentile??
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
assume body temp of healthy adults are normally distributed with a mean of 98.20F and a standard deviation of 0.62F. what body temperature is the 95th and 5th percentile??
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Find the z-value with a 95% left-tail: Ans: z = 1.645
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Find the z-value with a 5% left-tail: Ans: z = -1.645
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95%ile = 1.645*0.62+98.2 = 99.22 degrees
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5%ile = -1.645*0.62+98.2 = 97.18 degrees
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Cheers,
Stan H.
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