# Thread: probabilities- Please more detail that will easy for me to understand

1. ## probabilities- Please more detail that will easy for me to understand

• Assume random variable x is from a normal distribution with mean µ = 20 and standard deviation σ = 5, find the following probabilities:

a. P(14.735 £ x £ 30.325)
b. P(9.225 £ x £ 35)
c. P(x £ 27.5)
d. P(25 £ x £ 29.175)

• Given the profit payoff table below:
Decision alternative-d1 State of Nature s1

s1 s2 s3 s4

d1 12 9 10 4
d2 14 10 8 6
d3 9 12 15 11
d4 8 10 11 13

Suppose the manager has no probability information, what is the decision recommendation for him using the optimistic, conservative, and min-max regret approach?

2. in the part of calculating the probabilities, use the fact that:

P(a<x<b)=F(b)-F(a)

in your case, it's normal distribution, so what you have to do is to standarize every value:

z=(x-mu)/sigma
( z=(x-mean)/s.d)

and then just check the table of the standarized normal distribution

3. ## My answer- thanks for help :)

• P(14.735 £ x £ 30.325)
For value of 14.735
Z = (14.735 – 20)/5= -1.053
Probability from table = 0.3531

For value of 30.325
Z = (30.325 – 20)/5= 2.065
Probability from table = 0.4808
P(14.735 £ x £ 30.325) = 0.4808 + 0.3531 = 0.8339
• P(9.225 £ x £ 35)
For value of 9.225
Z = (9.225 – 20)/5= -2.155
Probability from table = 0.4846

For value of 35
Z = (35– 20)/5= 3
Probability from table=0.4986
P(9.225 £ x £ 35) = 0.4846 + 0.4986 = 0.9832
• P(x £ 27.5)
For value of 27.5
Z = (27.5– 20)/5 = 1.5
Probability from table = 0.4332
From table, for z=1.5, value is 0.4332
Probability P(x £ 27.5) = 0.5+ 0.4332 = 0.9332
• P(25 £ x £ 29.175)
For value of 25
Z = (25 – 20)/5= 1
Probability from table = 0.3413

For value of 29.175
Z = (29.175– 20)/5= 1.835
Probability from table = 0.4671
P(25 £ x £ 29.175) = 0.4671- 0.3413= 0.1258