Statistics Issues Analysis - Assignment Example

Question 2 (12 marks)

A bank is investigating the effectiveness of its small business loan application process and whether it should consider changing its practice. The current process consists of a loans officer making the decision to either “Approve” or “Not Approve” the loan application, based upon their subjective interpretation of certain criteria. As part of its investigation and for a trial period, the bank had a small set of summary information from each application entered into an automated “intelligent system”. The intelligent system then provided a risk rating for the application of either “Tolerable” (indicating the application was worthy of approval) or “Not Tolerable” (indicating the application was not worthy of approval). The applications were then assessed by loans officers in the usual manner, without knowledge of the prior risk rating.

It transpired that the intelligent system deemed 70% of applications to have “Tolerable” risk, with 60% of these subsequently being approved by the loans officers. 10% of applications that had been deemed “Not Tolerable” were also approved by loans officers.

• If T is used to represent the event “An application was deemed to have tolerable risk by the intelligent system” and A is used to represent the event “An application is approved by a loans officer”, then define in words the complement of both T and A. (and use this terminology in your working to answer the following questions)

Solution

• The complement of T (TC) is the event where “An application was NOT deemed to have tolerable risk by the intelligent system”.
• The complement of A (AC) is the event where “An application is NOT approved by a loans officer”.
• Draw a fully labelled tree diagram representing this situation, with the outcomes and probabilities shown along each branch and joint probabilities labelled and calculated at the end of each branch.

Solution

Tree diagram representing the situation

• Calculate the proportion of loan applications that were actually approved by loans officers. Compare this to the proportion the intelligent system would have approved – is it greater or less?

Solution

Thus, 45% of loan applications were actually approved by loans officers.

Proportion that the intelligent system would have approved is 70%.

The 45% proportion of loan applications that were actually approved by loans officers is less than the Proportion that the intelligent system would have approved.

• If a particular loan was approved by a loans officer, what is the probability that the intelligent system would not have approved it?

Solution

If a particular loan was approved by a loans officer, the probability that the intelligent system would not have approved it is 0.1 (proportion of loans approved by loan officers having rejected by the intelligent system).

• In probability terms, is “decision by the intelligent system” independent of the “loan officers’ decision”? Explain – noting that ‘yes’ or ‘no’ without justification will earn 0 marks.

Solution

Decision by the intelligent system is independent on the loan officers’ decision. This is because the probability of the decision made by the intelligent system does not in any way affect the probability of the decision being made by the loan officers.

Intrinsically, the probability of the two independent events are mutually exclusive, in that, the occurrence of one event does not affect the occurrence of the other.

Question 3 (8 marks)

A manufacturer producing carpets knows that its process producing a particular high end but complex carpet weave results in an average of 4.5 minor imperfections for each 10 linear metres of carpet.

• If the random variable in question here is labelled Y, identify the type of distribution Y has and write down the value(s) of its parameter(s).

Solution

Y follows a Poisson distribution. For this particular distribution, we have;

In this case;

•  is the average number of events per interval, which is
• e is the number 2.71828... (Euler's number) the base of the natural logarithms
• k takes values 0, 1, 2, …
• Determine the probability that a particular 10 metre section of this carpet chosen at random will have at most 6 minor imperfections.

Solution

After the calculations, we found that the probability that a particular 10 metre section of this carpet chosen at random will have at most 6 minor imperfections is 0.831051.

• A retailer is considering recommending this carpet type to a fussy homeowner who will need just 2 linear metres. Determine the probability that there will be at least 1 imperfection in this piece of carpet. Show your reasoning and the Excel formula used.

Solution

The excel formula used is;

=(EXP(-0.9)* (0.9)^0)/FACT(0)

Question 4 (16 marks)

Typically a small proportion of passengers booked on airline flights do not turn up to fly. Such passengers are called “no shows”. (For this reason some airlines actually overbook flights on frequent busy services in a strategy to limit the number of empty seats.)

• An airline’s historic data for a particular route serviced by their A34 plane which seats 52 passengers indicates that on average 4% of booked passengers are no shows.
• If we consider the number of “no shows” in this situation to be a random variable labelled X, identify the type of distribution X has and the value(s) of its parameter(s).

Solution

X follows a binomial distribution:

• Calculate the expected number of no shows on a fully booked flight on a particular A34 service and interpret this value in the context above.

Solution

This shows that the expected number of no shows on a fully booked flight on a particular A34 service is 2.08. That is to say, approximately 3 (considering persons) people will not show up on a fully booked flight on a particular A34 service.

• Calculate the probability of all passengers turning up for a fully booked flight on this service. Show the reasoning and the Excel formula used to make this calculation.

Solution

This value is close to zero

Excel formula;

=(FACT(52)/(FACT(52-52)*FACT(52)))*0.04^52*0.096^(52-52)

• Calculate the probability of having 4 or more no shows on a fully booked flight (i.e. 52 booked passengers) for this service. Show the reasoning and the Excel formula used to make this calculation

Solution

Excel formulas are;

=(FACT(52)/(FACT(52-0)*FACT(0)))*0.04^0*0.096^(52-0)

=(FACT(52)/(FACT(52-1)*FACT(1)))*0.04^1*0.096^(52-1)

=(FACT(52)/(FACT(52-2)*FACT(2)))*0.04^2*0.096^(52-2)

=(FACT(52)/(FACT(52-3)*FACT(3)))*0.04^3*0.096^(52-3)

• A regional airline services small communities in western New South Wales with planes that seat 10 passengers. Their historic data indicates a no show rate of 5%.
• Using the same approach as for part a. above, use statistical tables to find the probability of exactly 1 no show on a randomly selected flight that is fully booked.

Solution

Excel function;

=(FACT(10)/(FACT((10-1))*FACT(1)))*(0.05)^1*(0.95)^(10-1)

• You have used a particular probability distribution to answer question b. i). What assumption/s of this distribution is/are likely to be contravened in the context as described? Explain.

Solution

The most likely assumption to be contravened is the assumption that each trial is mutually exclusive, or independent of each other. This is because there are planes, and a no show in one plane may affect a no show in another depending on the factors such as the manner in which customers view the services offered. One disappointed client might fail to board other planes and also influence some other clients not to take the flights.

Question 5 (22 marks)

A company produces electric car batteries. The range of these batteries under standard driving conditions is understood to be normally distributed with a mean of 520 km and a standard deviation of 50 km.

• part (a)
• If the random variable described here is represented as X, then identify its type of distribution and write down the value(s) of its parameter(s).

Solution

X follows a normal distribution with mean, µ= 520 and standard deviation, = 50.

• Calculate the probability (using statistical tables) that a randomly selected battery will last less than 480 km under these standard driving conditions.

Solution

We seek to find P(x<480)

We obtain the z score as follows;

Thus, the probability that a randomly selected battery will last less than 480 km is 0.2119.

• Verify your answer to part ii) using the appropriate Excel statistical function and demonstrate you have done this by including the Excel formula used.

Solution

We also used excel formula. The formula used is given below;

=NORM.S.DIST(STANDARDIZE(480, 520, 50), TRUE)

=0.2119

• Calculate the probability (using statistical tables) that a randomly selected battery will last for more than 600 km under standard driving conditions.

Solution

We seek to find P(x>600)

We obtain the z score as follows;

Thus, the probability that a randomly selected battery will last for more than 600 km is 0.0548.

• Verify your answer to part iv) using Excel statistical functions and demonstrate you have done this by including the formula used.

Solution

=1-NORM.S.DIST(STANDARDIZE(600, 520, 50), TRUE)

=0.0548

• Use the NORMINV function to determine the range that just 2.5% of batteries would be expected to exceed under such standard driving conditions. Show the formula used and summarize your answer in a sentence.

Solution

The expected value is approximately 618 km. The formula used is;

=NORMINV((1-0.025), 520, 50)

The answer shows that given a probability of 0.025 (2.5%), a randomly selected battery will last for more than 618 km under standard driving conditions.

• The quality control officer for the company intends to test a random sample of 25 of these batteries under standard driving conditions.
• State the distribution of the mean distance travelled of such a sample.

Solution

The distribution of the mean distance travelled is 520 km. The mean distance is similar to the population mean.

• Calculate the probability that the average distance travelled would be between 510 km and 530 km and provide your answer in a sentence.

Solution

We seek to find P(510<x<530)

We obtain the z score as follows;

From the computations, we can see that the probability that the average distance travelled between 510 km and 530 km is 0.6826.