another statistics problem.?

Question

The webster mail order company sells expensive stereo by mail. The following table lists the frequency distribution of the number of orders received per day by this company during the past 100 days.

number of orders received per day 2 3 4 5 6
number of days 12 21 34 19 14

a. construct a probability distribution table for the number of orders received per day. Draw a graph of the probablility distribution.
b. is the probabilities listed in the table of part a. exact or approximate probabilities of various outcomes? Explain.
c. Let x= the number of orders recieved on any given day. Find the following probabilities
1. P(x=3)
2. P(x is > or = 3)
3. P(2 is < or = to x is < or = to 4)
4. P(x<4)

Answer 1

Dear Stacy,

This is not a well constructed problem, but that's probably not your fault. Part (a) depends on what distribution you decide upon. For instance, you could use something like a log-normal distribution and use the data to estimate its parameters. That said, let's be simplistic and say that the probabilities can be approximated by maximum likelihood, so the frequencies of the observations will represent our best guess of the probabilities. (Understand the problem with this, in that it effectively asserts that there is zero chance of receiving fewer than two orders or more than six, which could get you into big trouble if you were making decisions based on this approximation.)


Thus for (a):

graph & probability estimates
< | 0.00 (i.e., < 0 orders)
0 | 0.00
1 | 0.00
2 | * * * * * * * * * * * * 0.12
3 | * * * * * * * * * * * * * * * * * * * * * 0.21
4 | * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 0.34
5 | * * * * * * * * * * * * * * * * * * * 0.19
6 | * * * * * * * * * * * * * * 0.14
7 | 0.00
> | 0.00 (i.e., > 7 orders)


(b) Approximate. Some explanation was given above. Essentially we've asserted that what we saw in the past completely determines the distribution of outcomes of what we would expect to see in the future. This is quite a leap to make, given we have only observed 100 days.


(c)
(1) P(x=3) = 0.21 .

(2) P(x>=3) = P(x=3) + P(x=4) + P(x=5) + P(x=6) + P(x=7) + P(x>7)
= 1 - (P(x<0) + P(x=0) + P(x=1) + P(x=2))
= 0.88 .

(3) P(2<=x<=4) = P(x=2) + P(x=3) + P(x=4)
= 0.12 + 0.21 + 0.34
= 0.67 .

(4) P(x<4) = P(x<0) + P(x=0) + P(x=1) + P(x=2) + P(x=3)
= 0.33 .

Answer 2

sorry I don't know the answer to your question.