Using Python to answer these questions for NBC problem. And attached bellow the same data as the pervious answered questions

NBC problem

The file Bank.csvcontains data on 5000 customers of Universal Bank. The data include customer demographic information (age, income, etc.), the customer's relationship with the bank (mortgage, securities account, etc.), and the customer response to the last personal loan campaign

(Personal Loan). Among these 5000 customers, only 480 (=9.6%) accepted the personal loan that was offered to them in the earlier campaign. In this exercise, we focus on two predictors: Online (whether or not the customer is an active user of online banking services) and Credit Card (abbreviated CC below) (does the customer hold credit card issued by the bank), and the outcome Personal Loan (abbreviated Loan below).

Partition the data into training (60%) and validation (40%) sets.

a. Create pivot table for the training data with Online as column variable, CC as row variable, and Loan as a secondary row variable. The values inside the table should convey the count. Use the pandas dataframe methods melt()and pivot().

b. Consider the task of classifying customer who owns bank credit card and is actively using online banking services. Looking at the pivot table, what is the probability that this customer will accept the loan offer? [This is the probability of loan acceptance (Loan = 1) conditional on having bank credit card (CC = 1) and being an active user of online banking services (Online = 1).]

c. Create two separate pivot tables for the training data. One will have Loan (rows) as a function of Online (columns) and the other will have Loan (rows) as a function of CC.

d. Compute the following quantities [P(AB) means "the probability of A givenB"]:

1. P(CC = 1Loan = 1) (the proportion of credit card holders among the loan acceptors)
2. P(Online = 1Loan = 1)
3. P(Loan = 1) (the proportion of loan acceptors)
4. P(CC = 1Loan = 0)
5. P(Online = 1Loan = 0)
6. P(Loan = 0)

e. Use the quantities computed above to compute the naive Bayes probability

P(Loan = 1CC = 1, Online = 1).

f. Compare this value with the one obtained from the pivot table in (b). Which is a more accurate estimate?

g. Which of the entries in this table are needed for computing P(Loan = 1CC = 1, Online = 1)? In Python, run naive Bayes on the data. Examine the model output on training data, and find the entry that corresponds to P(Loan = 1 CC = 1, Online = 1). Compare this to the number you obtained in (e).

Here is the data:

ID	Age	Experience	Income	ZIP Code	Family	CCAvg	Education	Mortgage	Personal Loan	Securities Account	CD Account	Online	CreditCard
1	25	1	49	91107	4	1.6	1	0	0	1	0	0	0
2	45	19	34	90089	3	1.5	1	0	0	1	0	0	0
3	39	15	11	94720	1	1	1	0	0	0	0	0	0
4	35	9	100	94112	1	2.7	2	0	0	0	0	0	0
5	35	8	45	91330	4	1	2	0	0	0	0	0	1
6	37	13	29	92121	4	0.4	2	155	0	0	0	1	0
7	53	27	72	91711	2	1.5	2	0	0	0	0	1	0
8	50	24	22	93943	1	0.3	3	0	0	0	0	0	1
9	35	10	81	90089	3	0.6	2	104	0	0	0	1	0
10	34	9	180	93023	1	8.9	3	0	1	0	0	0	0
11	65	39	105	94710	4	2.4	3	0	0	0	0	0	0
12	29	5	45	90277	3	0.1	2	0	0	0	0	1	0
13	48	23	114	93106	2	3.8	3	0	0	1	0	0	0
14	59	32	40	94920	4	2.5	2	0	0	0	0	1	0
15	67	41	112	91741	1	2	1	0	0	1	0	0	0
16	60	30	22	95054	1	1.5	3	0	0	0	0	1	1
17	38	14	130	95010	4	4.7	3	134	1	0	0	0	0
18	42	18	81	94305	4	2.4	1	0	0	0	0	0	0
19	46	21	193	91604	2	8.1	3	0	1	0	0	0	0
20	55	28	21	94720	1	0.5	2	0	0	1	0	0	1
21	56	31	25	94015	4	0.9	2	111	0	0	0	1	0
22	57	27	63	90095	3	2	3	0	0	0	0	1	0
23	29	5	62	90277	1	1.2	1	260	0	0	0	1	0
24	44	18	43	91320	2	0.7	1	163	0	1	0	0	0
25	36	11	152	95521	2	3.9	1	159	0	0	0	0	1
26	43	19	29	94305	3	0.5	1	97	0	0	0	1	0
27	40	16	83	95064	4	0.2	3	0	0	0	0	0	0
28	46	20	158	90064	1	2.4	1	0	0	0	0	1	1
29	56	30	48	94539	1	2.2	3	0	0	0	0	1	1
30	38	13	119	94104	1	3.3	2	0	1	0	1	1	1
31	59	35	35	93106	1	1.2	3	122	0	0	0	1	0
32	40	16	29	94117	1	2	2	0	0	0	0	1	0
33	53	28	41	94801	2	0.6	3	193	0	0	0	0	0
34	30	6	18	91330	3	0.9	3	0	0	0	0	0	0
35	31	5	50	94035	4	1.8	3	0	0	0	0	1	0
36	48	24	81	92647	3	0.7	1	0	0	0	0	0	0
37	59	35	121	94720	1	2.9	1	0	0	0	0	0	1
38	51	25	71	95814	1	1.4	3	198	0	0	0	0	0
39	42	18	141	94114	3	5	3	0	1	1	1	1	0
40	38	13	80	94115	4	0.7	3	285	0	0	0	1	0
41	57	32	84	92672	3	1.6	3	0	0	1	0	0	0
42	34	9	60	94122	3	2.3	1	0	0	0	0	0	0
43	32	7	132	90019	4	1.1	2	412	1	0	0	1	0