Question

Can you help me with the binomial probability distribution? Please see instructions in the two images. Thank you.

Binomial Probability Distribution 1. Explain in your own words the importance of studying the binomial probability distribution. 2. Next, the probability distribution of a random variable is presented x. 20 25 30 35 f(x) 0.38 0.10 0.15 0.37 a) What is presented in the table, is it a probability distribution? Explain. b) What is the probability that x = 35? c) What is the probability that x is less than or equal to 25? d) What is the probability that x is greater than 30? 3. Next, a probability distribution is presented. X Probability f 0 2 0.10 0.15 ? 0.20 0.15 0.10 in 4 5 a) Find P (2) b) What is the expected (average) value of the number of services? c) What is the variance of the number of services? d) What is the standard deviation? 4. The probability distribution of the ratings established by employees on the relationship they have with their supervisors' ranges from 1 (poor) to 5 (excellent). Qualification of Supervisor Supervisor B the employee's A f(x) relationship with f(x) the supervisor (x) 0.05 0.04 2 0.09 0.10 0.03 0.12 4 0.42 0.46 0.41 0.28 ديا الرا a) What is the expected (mean) value of the securities offered to Supervisor A's rating? b) What is the expected (mean) value of the securities offered to Supervisor B's rating? c) Calculate the variance of the values offered to the rating of supervisor A. d) Calculate the standard deviation of the values offered for Supervisor B's rating. e) Compare satisfaction with Supervisor A with satisfaction with Supervisor B.

Answer 1

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).  $P(X \le 25)=P(x=20)+P(X=25)=0.38+0.1=0.48$

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 $\sum x*p(x)$ . 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 $E\left ( X-E(X) \right )^2=\sum (x-2.45)^2P(x)$

(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=$\sqrt{variance}=\sqrt{2.0475}=1.4309$

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: $\sqrt{1.1785}=1.0856$

e). The average rating for supervisor is more than B but the Variance of the B's ratings is less compared to A.