Can you help me with the binomial probability distribution? Please see instructions in the two images. Thank you.
1. The Binomial concept is very common in real life. Many things in our day to day life results in binary outcome. For example sex of a baby, success/failure,positive or negative. Many of these properties can easily be studies with the help of a binomial distribution be it to study the distribution or testing certain proportion or estimating outcome based on win/loss etc.
2. The probability distribution is:
x | p(x) |
20 | 0.38 |
25 | 0.10 |
30 | 0.15 |
35 | 0.37 |
a). The distribution of the random variable X is presented here with the associated probabilities. It is a probability distribution since the total probability is 1.
b). P(X=35)=0.37
c).
d). P(X>30)=P(X=35)=0.37
3. The distribution function is:
x | p(x) |
0 | 0.10 |
1 | 0.15 |
2 | ? |
3 | 0.20 |
4 | 0.15 |
5 | 0.10 |
a). P(x=2)=1-P(0)+P(1)+p(3)+p(4)+p(5)
=1-0.7=0.3
b). The expected value is given by . We form a table to accomplish this:
x | p(x) | x*p(x |
0 | 0.10 | 0 |
1 | 0.15 | 0.15 |
2 | 0.30 | 0.6 |
3 | 0.20 | 0.6 |
4 | 0.15 | 0.6 |
5 | 0.10 | 0.5 |
Total | 2.45 |
The expected value =2.45.
c). Variance is given by
(x-E(X))^2 | p(x) | V |
6.0025 | 0.10 | 0.6003 |
2.1025 | 0.15 | 0.3154 |
0.2025 | 0.30 | 0.0608 |
0.3025 | 0.20 | 0.0605 |
2.4025 | 0.15 | 0.3604 |
6.5025 | 0.10 | 0.6503 |
Total | 2.0475 |
The variance of the number of services is 2.0475
d). The standard deviation is sd=
4.
The calculations for A's satisfaction levels is presented below:
x | A | x*p(x) | x-E(x) | (x-E(x))^2 | V(A) | x | B | x*p(x) | x-E(x) | (x-E(x))^2 | V(B) | |
1 | 0.05 | 0.05 | -3.05 | 9.3025 | 0.4651 | 1 | 0.04 | 0.04 | -3.05 | 9.3025 | 0.3721 | |
2 | 0.09 | 0.18 | -2.05 | 4.2025 | 0.3782 | 2 | 0.1 | 0.2 | -2.05 | 4.2025 | 0.4203 | |
3 | 0.03 | 0.09 | -1.05 | 1.1025 | 0.0331 | 3 | 0.12 | 0.36 | -1.05 | 1.1025 | 0.1323 | |
4 | 0.42 | 1.68 | -0.05 | 0.0025 | 0.0010 | 4 | 0.46 | 1.84 | -0.05 | 0.0025 | 0.0011 | |
5 | 0.41 | 2.05 | 0.95 | 0.9025 | 0.3700 | 5 | 0.28 | 1.4 | 0.95 | 0.9025 | 0.2527 | |
Total | 4.05 | 1.2475 | Total | 3.84 | 1.1785 |
a) Expected value of supervisor A's rating=4.05
b). Expected value of supervisor B's rating=3.84
c). The variance of rating for supervisor A=1.2475
d). Standard deviation of Supervisor B:
e). The average rating for supervisor is more than B but the Variance of the B's ratings is less compared to A.
Can you help me with the binomial probability distribution? Please see instructions in the two images....
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