# You have a retail-clothing store for fashion items. For simplicity, let us consider just one type of gown in this question. You have to purchase your inventory at the start of the season at \$60 per...

You have a retail-clothing store for fashion items. For simplicity, let us consider just one type of gown in this question. You have to purchase your inventory at the start of the season at \$60 per gown, and you plan to sell gowns at \$100. You have a 50% chance of selling more than 10 gowns.

i) If you have to throw out unsold gowns at the end of the season, how much inventory will you order?

ii) However there is an active post-season sales market where you can sell gowns for \$B (B stands for ‘bargain’ sales). Plot on a cumulative probability graph the number you should order for values of B from \$0 to \$50.

mathsworkmusic | (Level 2) Educator

Posted on

i) Assume the probability `p ` of selling a gown ` `is constant and that the number of gowns ordered at the start of the season is equal to `n ` .

We can assume that the probability of selling `X ` gowns is Binomial(`n `,`p `). If the probability of selling more than 10 gowns is 0.5 then we can say that

`P(X>=10) = 1- Sigma_(x=1)^10 p^x(1-p)^(n-x) = 0.5 `

which implies that

`Sigma_(x=1)^10 p^x(1-p)^(n-x) = 0.5 `

Now assume that the probability `p_0 ` of selling an individual gown is such that the expected profit is zero (the balance point or break even point). In this instance

`60(1-p_0) = 40p_0 `

giving  `p_0 = 0.6 `.  Only values of `p ` greater than this will result in a positive expected profit per gown, so that we require `n ` corresponding to the locus  `p>=p_0 ` .  This implies that

`Sigma_(x=1)^10 0.6^x0.4^(n-x) >= 0.5`

Using statistical tables this gives  `n geq 18 `   since  Pr(`X >=10 ` ) `>= 0.5 ` for a Binomial(18,0.6) distribution.

This value for the number of gowns needed to be bought in order for the expected profit to be positive is that obtained in the absence of any knowledge about the probability of selling a gown `p `. If `p ` is known to be higher than the break even probability (this is likely as the set cost of a product is usually made taking profit margins into account so that expected profit works out greater than zero), then `n ` could be allowed to be lower.  Eg if we known that  `p=0.7 `  then using statistical tables we require `n >=15 ` .

Sources: