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?

My answer- thanks for help :)

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

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

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

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