Question 1147828: 6.4 4/The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of 8.9 cm.
a. Find the probability that an individual distance is greater than 215.00 cm.
b. Find the probability that the mean for 25 randomly selected distances is greater than 200.70 cm.
c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z>(x-mean)/sd or 215-202.5 all divided by 8.9
z>12.5/8.9 or >+1.40
probability is 0.0808
for the next part, the numerator is (sample mean-200.70) the denominator is sd/sqrt(n) or 8.9/sqrt(25) or 1.78
z>(200.70-202.5)/1.78
or z>-1.8/1.78 or probability of 0.8438
If the original sample size is normal, the sample may be considered to be normal as well. Also, there is nothing magic about 30. For normally distributed populations, small samples work. For skewed, larger ones are needed.
the fraction is now z>12.5/1.78
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