 # Reliability of System Configurations: Statistics and Probability Questions Answers SECTION A

Ql. (a) Consider the two system configurations in Figure 1.1. Each system consisting of four independent components that are connected as shown in Figure 1.1 (a) and Figure 1.1 (b), respectively. Given each component is operable with the probability of 0.8. Figure 1.1

For the system illustrated in Figure 1.1 (a). there are two subsystems connected in parallel, each one containing two components. In order for the system to ftmction, at least one of the two parallel subsystems must work. Within each subsystem, the two components are connected in series. so a subsystem will work only if all components in the subsystem work.

The system illustrated in Figure 1.1 (b) is obtained by connecting ties across the column of junction from the system in Figure 1.1 (a). Now the system fails as soon as an entire column fails, and system works only if every column works.

Discuss and compare which of the two system configurations in Figure 1.1 is more reliable. (7 marks)

1. Both Lily and James are reluctant to do house chores so their mother has decided to play a game. Both of them are required to draw a piece of paper from a box that contains of x number of blank papc(s) and a single piece of paper that marked cross, alternatively. one at a time. without replacement of the paper. The game will be won by whoever happens to draw the paper that marked cross and the winner will have to do house chores.

Being a gentleman, James offers Lily the choice of whether she wants to start first or not. Lily is afraid that she might won the game in the first draw. On the other hand, if she draws a blank paper in the first round. James's chances to draw that piece of paper that marked cross are increased.

Advise Lily whether she should take the offer to draw first or not, in order to avoid doing house chores. Explain your answer with numerical justification. (8 marks) [Total: 15 marks]

Q2. (a) An investor would like to invest R1+.1 10000 into shares of companies A and B. Let t be the amount of iinggits invested into company A and the remaining (10000 —1) be the amount of iinggits invested into company B. The graph of the variance of diversified portfolio considering both shares of companies A and B are shown in Figure 2.1 (a) and Figure 2.1 (b), respectively. Figure 2.1 (b)

Given both of the portfolios have taken into consideration of the identical expected return and standard deviation per share for companies A and B, respectively. Compare and give an explicit discussion about the difference between Figure 2.1 (a) and Figure 2.1 (b). (7 marks)

02. (b) The time that it takes for a driver to react to the brake lights on a decelerating vehicle is critically important in helping to avoid rear-end collisions. A recent study suggests that reaction time for an in-traffic response to a brake signal from the standard brake lights can be modeled with a normal distribution with mean of 1.06 sec and standard deviation of 0.4 sec.

1. Find the probability that the reaction time is between 0.8 sec and 1.8 sec. (4 marks)
2. Find the point Value k that has the property that 5% of all drivers have a reaction time of this value or lower. (4 marks) [Total: 15 marks]

Section A

Que 1 a)

System 1.1 a)

Table given below shows all possible states of the system. IS_OPERATIONAL columns indicates whether system will be operational or not.

Probability of system being operational =

(0.2)^2*(0.8)^2 +

(0.2)*(0.8)^3     +

(0.2)*(0.8)^3     +

(0.2)^2*(0.8)^2 +

(0.2)*(0.8)^3     +

(0.2)*(0.8)^3     +

(0.8)^4

= 0.8704

System 1.1 b)

Here I am assuming an column works if at least one component of  the column is operational (Please let me know if this assumption is wrong as we need to know about this but its not mentioned in the problem clearly).

Table given below shows all possible states of the system. IS_OPERATIONAL columns indicates whether system will be operational or not.

Probability of system being operational =

(0.2)^2*(0.8)^2 +

(0.2)^2*(0.8)^2 +

(0.2)*(0.8)^3     +

(0.2)^2*(0.8)^2 +

(0.2)*(0.8)^3     +

(0.2)*(0.8)^3     +

(0.2)*(0.8)^3     +

(0.2)^2*(0.8)^2 +

(0.8)^4

=0.9216

So, probability of system in 1.1 b) to be operational is higher than system in 1.1 a). So system in 1.1 b) is more reliable.

System 1.1 a)

 index A B C D IS_OPERATIONAL 0 0 0 0 0 NO 1 0 0 0 1 NO 2 0 0 1 0 NO 3 0 0 1 1 YES 4 0 1 0 0 NO 5 0 1 0 1 NO 6 0 1 1 0 NO 7 0 1 1 1 YES 8 1 0 0 0 NO 9 1 0 0 1 NO 10 1 0 1 0 NO 11 1 0 1 1 YES 12 1 1 0 0 YES 13 1 1 0 1 YES 14 1 1 1 0 YES 15 1 1 1 1 YES

System 1.1 b)

 index E F G H IS_OPERATIONAL 0 0 0 0 0 NO 1 0 0 0 1 NO 2 0 0 1 0 NO 3 0 0 1 1 YES 4 0 1 0 0 NO 5 0 1 0 1 NO 6 0 1 1 0 YES 7 0 1 1 1 YES 8 1 0 0 0 NO 9 1 0 0 1 YES 10 1 0 1 0 NO 11 1 0 1 1 YES 12 1 1 0 0 YES 13 1 1 0 1 YES 14 1 1 1 0 YES 15 1 1 1 1 YES

Que 1 b)

Case 1: Lily starts first

Pr(‘Lily drawing cross marked paper in first trial’) = 1/x+1

Pr(‘Lily drawing cross marked paper in second trial’) = x/x+1∗x-1/x1/x-1= 1/x+1

Pr(‘Lily drawing cross marked paper in third trial’) =  x/x+1∗x-1/x∗x-2/x-1∗x-3/x-2∗1/x-3= 1/x+1

Case 2: Lily starts second

Pr(‘Lily drawing cross marked paper in first trial’) = x/x+1∗1/x= 1/x+1

Pr(‘Lily drawing cross marked paper in second trial’) = x/x+1∗x-1/x1/x-2/x-1∗1/x-2= 1/x+1

Pr(‘Lily drawing cross marked paper in third trial’) = x/x+1∗x-1/x∗x-2/x-1∗x-3/x-2∗x-4/x-3∗1/x-4= 1/x+1

We can see that probability of drawing crossed paper is always equal to  independent of who starts drawing the paper first. So I would recommend Lily to ask James to start to draw the paper, this will lead to James taking the risk first.

Que 2

Part a)

Plot on part a looks to have lesser risk of the portfolio. Because at any value of t, the risk of the portfolio is lower in plot a as compared to plot b. One more way to explain it is the area under the curve is equal to risk of the portfolio, and it is lower for plot a.

Part b)

1. Probability that the reaction time is between 0.8 sec and 1.8 sec = cdf(1.8) – cdf(0.8) = 0.9678 – 0.2578 = 0.71
2. If point value k that has the property that 5% of all drivers have a reaction time of this value or lower,

then cdf(k) = 0.05

Solving for k, we will get k = 0.402

• Section B

Que 3 Data preparation task done

Que 4

KPL vs Fuel capacity  KPL vs Weight

With the limited number of sample points, we can see that there is linear association between KPL-Fuel Capacity and KPL-Weight variables.

The strength of linearity between KPL-Fuel Capacity looks higher than KPL-Weight.

Que 5

 Correlation bewtween KPL and Fuel Capacity is -0.53515 Correlation bewtween KPL and Fuel Capacity is -0.27739

There is negative correlation between both the pairs. The Magnitude of correlation between KPL-Fuel Capacity is higher than KPL-Weight,  this verifies our guess in Question 4.

Que 6

• a) Fitted regression line

KPL =  (-0.003192392)Weight + 21.54142512

• b) Interpretation

The linear regression equation above says that for every 1 unit change in weight, there is a constant decrement of 0.003192392 in KPL.

• c) This model is significant at 5% significance level. P-value of coefficient of weight is 0.031 and P-value of intercept is much lesser than 0.01.
• d) Co-efficient of determination also known as R-square statistic for the model is = 0.07694. Value of R-square statistic ranges from 0 to 1. Higher the value of R-square better is the predictive power of the model. So for our current model R-square is now very good.
• e) When we fit linear regression line, our objective function to be optimised is squared differences of actual Y value and predicted Y value. That is why linear regression is also called as least square regression line.
• f)

KPL =  (-0.003192392)Weight + 21.54142512

Lower value of coefficient of weight at 95% significance level = -0.0060986

Lower value of intercept at 95% significance level = 16.2962836

Upper value of coefficient of weight at 95% significance level = -0.0002862

Upper value of intercept at 95% significance level = 26.7865666

So, 95% confidence interval of KPL =

[-0.0060986*2560 +16.2962836 , -0.0002862*2560 + 26.7865666]

[0.6838675999999992, 26.0538946]

Interpretation of this 95% confidence interval is that if we randomly change the training data, then 95% of the times, KPL for a car with weight 2560kg will lie between [0.6838675999999992, 26.0538946]

Que 7

To determine if average value of KPL for sedan is statistically equal to average value of KPL of hatchback, we will do unpaired t-test.

H0: Average value of KPL for sedan is statistically equal to average value of KPL of hatchback

Ha: Average value of KPL for sedan is statistically not equal to average value of KPL of hatchback

If the p-value of t-test < 0.05 we reject H0 and accept Ha. But if the p-value of t-test is > 0.05 then we accept H0.

In our case p-value = 0.0005, so we reject H0.

So, average value of KPL for sedan is statistically not equal to average value of KPL of hatchback.