Question 974217: Assume body temps of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F.
1. If you have a body temp of 99.0 F, what is your percentile score?
2. Convert 99.0 F to a standard score (or z score)
3. Is a body temp of 99.0 F unusual? Why or why not?
4. 50 adults are randomly selected. What is the likelihood that the mean of their body temps is 97.98 F or lower?
5. A person's body temp is found to be 101. F. Is the result unusual- why?
6. What body temp is in the 95th percentile?
7. What body type is in the 5th percentile?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! mean is 98.2;;;;;z=(x-mean)/sd ;; For a sample, (x-mean)/[sd/sqrt(n)]
sd is 0.62
1. z=(99-98.2)/0.62= +1.29 90th percentile (slightly more)
2. see 1.
3. No. I am a retired physician, and this may be found for a variety of reasons. It can be normal, but in of itself, it is usually ignored. Statistical and clinical significance are two different entities.
4.z=(-0.22)*sqrt(50)/0.62 the SE is 0.62/sqrt(50). I inverted to divide and put the sqrt in the numerator. It is the same thing. z=-2.52;; Probability of this being random is 0.006.
5. Yes. It is more than 4 standard deviations above the mean. This is highly unlikely to be normal.
6. z=1.645 so 0.62*1.645 + Mean =99.22
(x-mean)/sd =1.645, the 95th percentile. It is +1.02 degrees.
7.It is the same amount in the other direction, - 1.02 degrees or 97.18. Note that for one person, a low temperature is possible whereas for a sample of 50, even 20% of this decrease would be considered significant.
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