# In a distribution exactly normal 7% of the items are under 35 and 89% are under 63,what the mean and standard deviation of the distribution?

The Z value corresponding to a probability of 7% is found from a suitable table or on-line site. It is -1.4758. Similarly, the Z value corresponding to a probability of 89% is found to be 1.2265. If "s" is used to represent the standard deviation, the difference between 35 and 63 represents 1.2265-(-1.4758)=2.7023 times s. Thus
s = (63-35)/2.7023 ≈ 10.362

At this point, we know that 63 is 1.2265*s above the mean, so the mean (m) must be
m = 63 - 1.2265*10.362 = 50.292

Given that the original problem statement uses numbers with 2 significant figures, the answer should be expressed to 2 significant figures.

The mean and standard deviation of the distribution are 50 and 10, respectively.
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A check of the probability of 35 in a normal distribution of mean 50 and standard deviation of 10 gives 6.7%, which rounds to 7%. However, the same check for 63 gives 90.3%, a bit higher than the value given in the problem. Thus, you may need to use (m, s) = (50.3, 10.4) to make the numbers work out, even though this level of precision is not really supported by the numbers in the problem.
thanked the writer. 