The average demand for rental skis on winter Saturdays at a particular area is 150 pairs, which has been quite stable over time. There is variation due to weather conditions and competing areas; the standard deviation is 20 pairs. The demand distribution seems to be roughly normal.

(a) The rental shop stocks 170 pairs of skis. What is the probability that demand will exceed this supply on any winter Saturday?
(b) How many pairs of skis in stock does the shop have to have to make the probability in question (a) less than .01?

a) 15.87% probability that demand will exceed this supply on any winter Saturday

b) The shop needs to have 197 pairs of skis in stock.

Step-by-step 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.

In this problem, we have that:

(a) The rental shop stocks 170 pairs of skis. What is the probability that demand will exceed this supply on any winter Saturday?

This probability is 1 subtracted by the pvalue of Z when X = 150. So

has a pvalue of 0.8413

1 - 0.8413 = 0.1587

15.87% probability that demand will exceed this supply on any winter Saturday

(b) How many pairs of skis in stock does the shop have to have to make the probability in question (a) less than .01?

This is X when Z has a pvalue of 0.99.

So X when Z = 2.33

Rounding up, since the number of pairs is a discrete number

The shop needs to have 197 pairs of skis in stock.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204