Probability Distribution: Own more credit cards, more purchases with them?

Probability Distribution

It is a case study - "Own more credit cards, more purchases with them?"


Qns 1) Do owning more cards cause one to buy more items with them?
Qns 2) Find mean of the number of cards owned per person
Qns 3) Find the correlation and covariance

Koliex said:

It is a case study - "Own more credit cards, more purchases with them?"

Number of cards	Cards owned	Buying Frequency/week
0	0	1
1	5	4
2	15	12
3	6	6
>=4	5	5

Qns 1) Do owning more cards cause one to buy more items with them?
Qns 2) Find mean of the number of cards owned per person
Qns 3) Find the correlation and covariance

Click to expand...

What is the difference between "number of cards" and "cards owned"?

Question 2 asks for "per person" information, but I don't see any "number of people" info in the table...?

When you reply, please include a clear listing of your thoughts and efforts so far, even if you think it's wrong. Thank you!

stapel said:

What is the difference between "number of cards" and "cards owned"?

Question 2 asks for "per person" information, but I don't see any "number of people" info in the table...?

When you reply, please include a clear listing of your thoughts and efforts so far, even if you think it's wrong. Thank you!

Click to expand...

Hi, thank you for replying!

The number of people was supposed to be 31 but after collecting the data, it appears that the total does not tally 31 for purchase frequency. I was supposed to come up with some assumptions to explain why the other 4 people data were not reflected in the table. For the assumption, I thought I could manipulate the data by categorising those 4 people under the ones who did not purchase anything in a week. This is done so I can get total of 31 for Buying Frequency/week.

I interpreted that perhaps the first and second column are interrelated such as for e.g. Num of cards: 1 and cards owned: 5 meant that there are 5 people who owned 1 card each.

I tried attempting the question in the following way:


1) What is the probability of purchasing more given that one owns more card
P(own more than 4 cards and purchases made) = 5/31 * 5/31 = 0.021
P(own 0 card and purchases made) = 0 * 4/31 = 0
P(own 1 card and purchases made) = 5/31 * 12/31 = 0.021
P(own 2 card and purchases made) = 4/31 * 4/31 = 0.187
P(own 3 card and purchases made) = 6/31 * 6/31 = 0.037

2) The mean of the num of cards per person:
Mean = ((0*0)+(1*(5/31))+(2*(5/31))+(3*(6/31))+(1-((0*0)+(1*(5/31))+(2*(5/31))+(3*(6/31))) = 6.12

I think the way I am attempting the problem is incorrect but I have no idea how to tackle this case study