Answer:


Explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Middle 85%
Goes from X when Z has a pvalue of 0.5 - 0.85/2 = 0.075 to X when Z has a pvalue 0.5 + 0.85/2 = 0.925.
Lower limit is 8
This means that when X = 8, Z has a pvalue of 0.075. So when
.




Upper limit is 14
This means that when X = 14, Z has a pvalue of 0.925. So when
.



From the lower limit, we have that

So





And
