Asked by lauren
An important indicator of lung function is forced expiratory volume(FEV), which is the volume of air that a person can expire in one second. Dr. Jones plans to measure FEv in a random sample of n young women from a certain population, and to use the sample mean as an estimate of the population mean. Let E be the event that Jones's sample mean will be within +- 100mLi of the population mean.Assume that the population distribution is normal with mean 3000 mLi and standard deviation 400 mLi. Find Pr{E} if:
A)n=15
A)n=15
Answers
Answered by
Damon
mean of sample means = mean of population = 3000
sigma sample means = sigma population/sqrt n
so s sample mean = 400/sqrt(15)
= 103
100/103 = .971 s
so we are interested in between the mean -.971 sigma to the mean +.971 sigma
from tables of normal distribution you can get the probability of being beyond mean +.971 sigma
it is 1-.834 = .166
by symmetry the probability of being below mean -.971 sigma is also .166
so the probability of being above or below the desired range i 2*.166 = .332
so the probability of being within the desired range is
1-.332 = .668 or 67 %
This makes sense because we all know that about 68 % lies within 1 sigma and we are at almost 1 sigma, namely .971 sigma
sigma sample means = sigma population/sqrt n
so s sample mean = 400/sqrt(15)
= 103
100/103 = .971 s
so we are interested in between the mean -.971 sigma to the mean +.971 sigma
from tables of normal distribution you can get the probability of being beyond mean +.971 sigma
it is 1-.834 = .166
by symmetry the probability of being below mean -.971 sigma is also .166
so the probability of being above or below the desired range i 2*.166 = .332
so the probability of being within the desired range is
1-.332 = .668 or 67 %
This makes sense because we all know that about 68 % lies within 1 sigma and we are at almost 1 sigma, namely .971 sigma
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