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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 $X\sim\text{Poi}$ 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. By the fundamental calculation this is the same as $\mathbb{P}[z_1 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 $=0.6554-\mathbb{P}[z>0.93]=0.6554-\mathbb{P}[\text{not-}z\leq 0.93]$ $=0.6554-(1-\mathbb{P}[z\leq 0.93])=0.6554-(1-0.8238)=0.4792$.

## 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