Question 919955: THE AVERAGE AGE OF C.E.Os, IS 56 YEARS. ASSUME THE VARIABLE IS NORMALLY DISTRIBUTED. IF THE STANDARD DEVIATION IS 4 YEARS, FIND THE YOUNGEST AND THE OLDEST PERSON WHO QUALIFIES TO BE A C.E.O.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you cannot absolutely find the youngest person or the oldest person who qualifies to be a ceo using the normal distribution.
you can only find the probability that the youngest person will be less than a certain age and the probability that the oldest person will be greater than a certain age.
what you need to be asking is something like:
what is the youngest possible age and oldest possible age so that the probability that a person can be ceo of a corporation is 99% or some other figure, assuming the average age is 56 years old and the standard deviation is 4 years.
now you have something you can look up in a table or use a calculator to find.
for example:
at 99% confidence interval, the z-score will be between -2.576 and 2.576 rounded up to 3 decimal places.
the probability of a z-score being between -2.576 and 2.576 is at least 99%.
now you want to translate the z-score to a raw score.
that is done as follows:
the formula is:
z = (x-m)/s
z is the z-score.
x is the raw score
m is the mean
s is the standard deviation.
in your problem:
z = plus or minus 2.576.
x = what you want to find.
m = the average age of a ceo which is 56 years of age.
s is the standard deviation which is 4 years.
you want to solve for x.
start with:
z = (x-m)/s
multiply both sides of this equation by s to get:
z*s = x-m
add m to both sides of this equation to get:
z*s + m = x which is the same as:
x = z*s + m
replace z with 2.576 and s with 4 and m with 56 and you get:
2.576 * 4 + 56 = 66.304 which you can round up to 67.
replace z with -2.576 and s with 4 and m with 56 and you get:
-2.576 * 4 + 56 = 45.696 which you can round down to 45.
rounding the low score down and the high score up is a conservative estimate that ensures you will be at least 99% confident the age will be between those limits.
there is at least a 99% probability that the age of a ceo will be between 45 and 67 years of age.
that still leaves at most 1% of ceos that could either be above 67 or below 45 years of age.
if you want to be more certain, then raise the confidence interval.
assume you want to be 99.999% sure that the age of the ceo will be within certain limits.
that generates a different z-score.
99.999% probability generates a z-score of plus or minus 4.418 rounded up to 3 decimal places.
the same formula applies.
x = z*s + m
on the high side, you will get x = 4.418 * 4 + 56 = 73.672 which you can round up to 74.
on the low side, you will get x = -4.418 * 4 + 56 = 38.328 which you can round down to 38.
there is at least 99.999% probability that a ceo will be between the ages of 38 and 74.
that doesn't mean there can't be a ceo younger than 38 years of age and doesn't mean that there can't be a ceo older than 74 years of age.
it just means that the probability of that happening is very low.
in fact, the probability that a ceo will be less than 38 years of age or more than 74 years of age is at most .001%
if you want to see this visually, the following pictures might help.
99% probability:
99.999% probability:
the numbers are slightly different due to rounding errors and the fact that i took a conservative view of the z-score required while the calculator simply rounded up or down depending on whether the next digit was 5 or more for up and less than 5 for down.
round the low numbers shown in the picture down to the nearest integer and the high number shown in the picture up to the nearest integer and we agree.
for 99%, i got 45 to 67 and the picture shows 45 to 67.
for 99.999%, i got 38 to 74 and the picture shows 38 to 74.
any differences would have been due to rounding errors and assumptions.
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