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Test 2 A Probability Solutions

The CFO of a large civil engineering company is concerned about the number of sick days taken by its 2,000 construction workers and hires a business analyst to monitor the situation.

Question 1(a)

Historic records suggest that workers take an average of 5 sick days out of 220. The analyst assumes that the number of days missed in a year of 220 days has a Binomial Distribution. Calculate the probability that

Let X be the number of sick days that a worker takes in a year. The question says that  X is binomially distributed. The number of trials,  n, is 220 and the probability of a sick day… well if it is 5 out of 220 then it is  \displaystyle \frac{5}{220}=\frac{1}{44} and we have that  X\sim \text{Bin}[220,1/44].

i. a worker takes three or more sick days in a year

We are looking for for the probability that X is three or more

\mathbb{P}[X\geq 3]=\mathbb{P}[\text{not-}(X=0\text{ or }X=1\text{ or }X=2)]

=1-\mathbb{P}[X=X=0\text{ or }X=1\text{ or }X=2]

=1-(\mathbb{P}[X=0]+\mathbb{P}[X=1]+\mathbb{P}[X=2])

\displaystyle 1-\left({220\choose 0}\left(\frac{1}{44}\right)^0\left(\frac{43}{44}\right)^{220}+{220\choose 1}\left(\frac{1}{44}\right)^1\left(\frac{43}{44}\right)^{219}\right.

\displaystyle \left.+{220\choose 2}\left(\frac{1}{44}\right)^2\left(\frac{43}{44}\right)^{218}\right)

=1-(0.0063602+0.032541+0.08286)=0.8782

ii. a worker takes no sick days in a year.

We are looking for the probability that X is zero. We have actually found this above

\displaystyle \mathbb{P}[X=0]=0.00636.

Question 1 (b)

If the probability that a worker takes a sick day is \frac{5}{220}, on average how many sick days do you expect on a given day on one typical job which has 66 workers.

Well \frac{5}{220}=\frac{1}{44} of the workforce: \displaystyle \frac{1}{44}\times 66=1.5 per day.

 The analyst believes that the daily number of workers out sick has a Poisson Distribution. Calculate the probability that no sick days are taken on the job on two days in a row.

We have from the previous that the mean daily number of workers out sick is $\lambda=1.5$.   However over two days the mean number of sick days is 2\times 1.5=3. Let X be the daily number of  workers out sick over two given days. We have that X\sim\text{Poi}[3] and

\displaystyle \mathbb{P}[X=0]=\frac{e^{-3}(3)^0}{0!}=e^{-3}=\frac{1}{e^3}.

Question 1 (c)

Total absenteeism across the workforce has become an issue with shareholders. The analyst believes that the total number of sick days taken per day is normally distributed with a mean of 57 and a standard deviation of 7.5. Calculate the probability that

there are more than 72 sick days taken on a given day.

Let X be the total number of sick days per day. We have that X\sim N[57,7.5] and we are looking for \mathbb{P}[X> 72]. By the fundamental calculation:

\mathbb{P}[X> 72]=\mathbb{P}[z>z_1],

where z\sim N[0,1] and z_1 is the z-transform of X_1=72:

\displaystyle z_1=\frac{X_1-\mu}{\sigma}=\frac{72-57}{7.5}=2.

Now we have

\mathbb{P}[X>72]=\mathbb{P}[z>2]=\mathbb{P}[\text{not-}(z\leq 2)]

=1-\mathbb{P}[z\leq 2]=1-0.9772=0.0228.

 there are between 50 and 60 sick days taken on a given day.

Now we are looking for \mathbb{P}[50<X<60]. By the fundamental calculation this is the same as \mathbb{P}[z_1<z<z_2] where

\displaystyle z_1=\frac{50-57}{7.5}\approx -0.93 and \displaystyle z_2=\frac{60-57}{7.5}=0.4.

Now, possibly using areas, and definitely symmetry, we make the calculation

\mathbb{P}[-0.93<z<0.4]=\mathbb{P}[z<0.4]-\mathbb{P}[z<-0.93]

=0.6554-\mathbb{P}[z>0.93]=0.6554-\mathbb{P}[\text{not-}z\leq 0.93]

=0.6554-(1-\mathbb{P}[z\leq 0.93])=0.6554-(1-0.8238)=0.4792.

Course Notes

Finally completed: here.

Week 11

In Week 11 we looked at Sampling Theory.

Week 12

In Week 12 we will finish the module by looking at Hypothesis Testing and Control Charts.

Week 13: Review Week

I will be available to any and all students (Groups A & B) at the following (usual) times and (usual) venues:

  • Review Lecture Monday 09:00 B212
  • Review Lecture Tuesday 14:00 B212
  • Review Lecture Wednesday 11:00 B212
  • Review Tutorial Thursday 15:00 B228

The Review Lectures will be conducted as follows (from Monday 9 December)

  1. Students can ask any question and I will answer it on the whiteboard. If we run out of questions
  2. I will start going through the Autumn 2013 paper (which was given out in the Thursday 28 December tutorial). If we finish this paper
  3. I will help ye one to one.

The Review Tutorials will be conducted as follows

  1. Students can ask any question and I will answer it on the whiteboard. If we run out of questions
  2. I will help ye one to one

Academic Learning Centre

I would urge anyone having any problems with material that isn’t being addressed in the tutorials to use the Academic Learning Centre. As you can see the timetable is quite generous. You will get best results if you come to the helpers there with specific questions.

Math.Stack Exchange

If you find yourself stuck and for some reason feel unable to ask me the question you could do worse than go to the excellent site math.stackexchange.com. If you are nice and polite, and show due deference to these principles you will find that your questions are answered promptly. For example What is a Confidence Interval?

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