Question on Normal Distribution

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sonikmook

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Hi everyone,
i just can’t seem to understand what formula or methodology to apply to this question:
The percentage changes in annual earnings for a company are approximately normally distributed with a mean of 5% and a standard deviation of 12%. The probability that the average change in earnings over the next five years will be greater than 15.5% is closest to 2.5%.
I suppose the standard error can be calculated by 12/square root of 5. However, even then, i don’t understand why the mean should be in the denomintor. Isn’t it usually “n”?
Thanks!
 
The standard error is 12%/√5. (Don’t forget the percent sign.)
I don’t understand your comment about the denominator: the mean (5%) isn’t in the denominator; the number of years (5) is in the denominator.
 
oh yes ur right…the denominator is the number of years.
I don’t understand how to get the answer of 2.5% though. I’m not really sure what formula to apply for this.
 
P(X>15.5) = 1 - P(X<15.5)
P(X<15.5) = P(Z< (15.5 - mu)/std err))
So calculate the z-value which is
z = (15.5-5)/(12/sqrt(5))
Look up the value of z in the standard normal distribution table.
Your answer is 1 minus the above mentioned probability.
 
z = 1.956 - you should have this committed to memory as +/- 1.96 being the reliability factors (z-critical values) for a 95% confidence level– that is, the probability between these cutoffs is 0.95.
Since you are looking at only the upper tail, divide the total tail probability by two—> (1-0.95)/2 = 0.05/2 = 0.025 or 2.5%…
 
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