In: Statistics and Probability
Let X~BIN(40,0.1). Let Y=max(0,x-3). FindE[Y].
Since X is binomially distributed over (40,0.1), i calculated the distribution using calculation and table is given below:
Binomial distribution (n=40, p=0.1) | |||
f(x) | |||
x | Pr[X = x] | ||
0 | 0.0148 | ||
1 | 0.0657 | ||
2 | 0.1423 | ||
3 | 0.2003 | ||
4 | 0.2059 | ||
5 | 0.1647 | ||
6 | 0.1068 | ||
7 | 0.0576 | ||
8 | 0.0264 | ||
9 | 0.0104 | ||
10 | 0.0036 | ||
11 | 0.0011 | ||
12 | 0.0003 | ||
13 | 0.0001 | ||
14 | 0 | ||
15 | 0 | ||
16 | 0 | ||
17 | 0 | ||
18 | 0 | ||
19 | 0 | ||
20 | 0 | ||
21 | 0 | ||
22 | 0 | ||
23 | 0 | ||
24 | 0 | ||
25 | 0 | ||
26 | 0 | ||
27 | 0 | ||
28 | 0 | ||
29 | 0 | ||
30 | 0 | ||
31 | 0 | ||
32 | 0 | ||
33 | 0 | ||
34 | 0 | ||
35 | 0 | ||
36 | 0 | ||
37 | 0 | ||
38 | 0 | ||
39 | 0 | ||
40 | 0 |
Now Y = max (0, x-3).
This means that,
Y = 0, for all x 3
Y = x-3. for all x 4
Y | x | P(x) | Y*P(x) |
0 | 0 | 0.0148 | 0 |
0 | 1 | 0.0657 | 0 |
0 | 2 | 0.1423 | 0 |
0 | 3 | 0.2003 | 0 |
1 | 4 | 0.2059 | 0.2059 |
2 | 5 | 0.1647 | 0.3294 |
3 | 6 | 0.1068 | 0.3204 |
4 | 7 | 0.0576 | 0.2304 |
5 | 8 | 0.0264 | 0.132 |
6 | 9 | 0.0104 | 0.0624 |
7 | 10 | 0.0036 | 0.0252 |
8 | 11 | 0.0011 | 0.0088 |
9 | 12 | 0.0003 | 0.0027 |
10 | 13 | 0.0001 | 0.001 |
11 | 14 | 0 | 0 |
12 | 15 | 0 | 0 |
13 | 16 | 0 | 0 |
14 | 17 | 0 | 0 |
15 | 18 | 0 | 0 |
16 | 19 | 0 | 0 |
17 | 20 | 0 | 0 |
18 | 21 | 0 | 0 |
19 | 22 | 0 | 0 |
20 | 23 | 0 | 0 |
21 | 24 | 0 | 0 |
22 | 25 | 0 | 0 |
23 | 26 | 0 | 0 |
24 | 27 | 0 | 0 |
25 | 28 | 0 | 0 |
26 | 29 | 0 | 0 |
27 | 30 | 0 | 0 |
28 | 31 | 0 | 0 |
29 | 32 | 0 | 0 |
30 | 33 | 0 | 0 |
31 | 34 | 0 | 0 |
32 | 35 | 0 | 0 |
33 | 36 | 0 | 0 |
34 | 37 | 0 | 0 |
35 | 38 | 0 | 0 |
36 | 39 | 0 | 0 |
37 | 40 | 0 | 0 |
Sum= | 1.3182 |
Now E[Y] = Y * P(x)
E[Y] = 1.3182