In this posts, you are going to find many solved exercises about probability used in operations.
Exponential Distribution
Exercise 1
In an industrial environment, the work accidents occur at a rate of 0.5 / week, in a five days week. Calculate:
- The probability of not occurring any accident in the first week.
- The probability of occurring one accident on Wednesday of the second week.
First, let's see the information that we have got: λ=0.1 (accidents per week) which comes from 1 accident divided by 10 days (5 day-week). Then, we have x=5 days. So we apply the formula in Excel: =DIST.EXPON(x;λ;1). Here, ‘1' belongs to cumulative.
P1: 1-[=DIST.EXPON(5;0,1;1)] = 1-0,3934 = 0,60653.
Second, to find out the probability on Wednesday, we need to evaluate the exponential distribution on Wednesday (T≤8) minus the previous days (T<7). Now, x=10 days.
P1: =DIST.EXPON(8;0,1;1)-DIST.EXPON(7;0,1;1) = 0,047.
Normal Distribution
Exercise 1
Use the excel sheet to calculate Pr[1.5 < X < 4.3], given that X ~ Gauss (3,2).
For this exercise, we need to just apply the formula in Excel, which is =DIST.NORMAL(x;µ;σ;1). The variables are: x1=4,3; x2=1,5; µ=3 (also called “mean”); σ=2 (also called “standard deviation”).
Pr: =DIST.NORMAL(4,3;3;2;1)-DIST.NORMAL(1,5;3;2;1) = 0,7421-0,2266 = 0,5155.
Statistical Inference
Exercise 1
Let x1,x2,…,x9 a random sample drawn from the population X~ Gauss(µ, 1). Test the hypothesis H0:μ=2 vs HA: μ≠2.
- In which case(s) you reject H0: x~ = 2.88.
- Calculate the p-value associated with each test, for alpha = 0.05.
Let's get first the T values using this formula: T=(“mean of x~”-µ)/(σ/(“square root of n”)). The data is: (mean of) x~=2,88, σ=1, µ=2, n=9. The limits are 0 and 1 for the Gaussian distribution.
T=(2,88-2)/(1/SQRT(9)) = 2,64.
To evaluate if it gets rejected, let's compare it with the limits of the graph: ±1,96. In this case, it is out of the ‘x' values admitted so it is rejected. Now, in order to get the p-values, let's apply the formula for normal distribution: =DIST.NORMAL(x;µ;σ;1) times 2 for a bilateral check.
P = (1 – [=DIST.NORMAL(2,64;2;1;1]) * 2= (1-0,7389)*2 = 0,26108
Since the argument for this test is that it must be lower than alpha=5%. Therefore, it fails to do so.
