# Homework Help: No. of customers (probability)

1. Mar 18, 2006

### Clari

The expenses of customers follows a normal distribution with mean equals to $428, and standard deviation equals to$100. A customer spending \$600 or more will receive a gift. If the probability of the shopping centre giving out gifts is at least 0.99, find the smallest no. of customers visiting the store.

My solution is something like:
X: expenses of customers
X~N(428, 100)
P(X>= 600)
= P(Z>= (600-428)/100 )
= P(Z>= 1.72)
= 0.5 - A(1.72)
From the table, P(X>= 600) = 0.5 - 0.4573= 0.0427

n: no. of customers
0.0427^n >= 0.99 <----Why isnt this expression incorrect?
There is another expression (1-0.0427)^n <= 0.01, which is correct. But as far as i'm concerned, they should be both correct Please help!

2. Mar 18, 2006

### 0rthodontist

You could try actually solving those equations for n to see which one makes sense. The first one says the upper limit on the number of customers is 0.

The way to figure it out is to think about what they are asking. They are looking for a number of customers n so that the probability of AT LEAST one customer receiving a gift is .99 or more. This is the same as saying that the probability that none of the customers receive gifts is .01 or less. And the probability that a given customer does not receive a gift is 1 - .0427, so you take that number to the nth to get the probability that all customers do not receive gifts.

What your first (incorrect) expression says is that the probability that ALL customers receive gifts is at least .99. Clearly you can't have very many customers for this to happen!

3. Mar 18, 2006

### Cyrus

$$\frac {\bar x - \mu} { \frac {\sigma} { \sqrt{n}}}= z$$

Find a z-score that corresponds to 0.99 and solve for n.

Last edited: Mar 18, 2006