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Which of the following random variables could be modelled with a binomial distribution and which could be modelled with a Poisson distribution?

You will lose 1 mark for every incorrect answer. The minimum mark is 0.
Binomial Distribution Poisson Distribution
A salesperson has a 50% chance of making a sale on a customer visit and she arranges 10 visits in a day. Let X be the number of sales that day.
We are interested in X, the number of machine breakdowns in a day. Such breakdowns at a particular IT company occur at a rate of eight per week.
On average, three patients arrive at a local Accident and Emergency department every hour. We count the number, X, of patients in an hour period.
Expected answer:
Binomial DistributionPoisson Distribution
A salesperson has a 50% chance of making a sale on a customer visit and she arranges 10 visits in a day. Let X be the number of sales that day.
We are interested in X, the number of machine breakdowns in a day. Such breakdowns at a particular IT company occur at a rate of eight per week.
On average, three patients arrive at a local Accident and Emergency department every hour. We count the number, X, of patients in an hour period.

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

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The mean number of sales per day at a telecommunications centre is $\var{thismany}$.

Employees receive a warning if they make less than $\var{number1}$ per day.

 

a)

Assuming a Poisson distribution for $X$, the number of sales per day, write down the value of $\lambda$.

$X \sim \operatorname{Poisson}(\lambda)$

$\lambda = $  Expected answer:

Find $\operatorname{E}[X]$ the expected the number of sales per day.

$\operatorname{E}[X]=$  Expected answer:

Find the standard deviation for daily sales..

Standard deviation = Expected answer: (to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Find the probability that a randomly selected employee makes exactly $\var{thisnumber}$ sales.

$\operatorname{P}(X=\var{thisnumber})=$  Expected answer: (to 3 decimal places).

 

Find the probability that a randomly selected employee receives a warning.

Probability = Expected answer: (to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

a)

1. $X \sim \operatorname{Poisson}(\var{thismany})$, so $\lambda = \var{thismany}$.

2. The expectation is given by $\operatorname{E}[X]=\lambda=\var{thismany}$

3. $\operatorname{stdev}(X)=\sqrt{\lambda}=\sqrt{\var{thismany}}=\var{sd}$ to 3 decimal places.

b)

1. \[ \begin{eqnarray*}\operatorname{P}(X = \var{thisnumber}) &=& \frac{e ^ { -\var{thismany}}\var{thismany} ^ {\var{thisnumber}}} {\var{thisnumber}!}\\& =& \var{prob1} \end{eqnarray*} \] to 3 decimal places.

2. If an employee receives a warning then he or she must have sold less than 2.

Hence we need to find:

\[ \begin{eqnarray*}\operatorname{P}(X < \var{number1})& =& \simplify[all,!collectNumbers]{P(X = 0) + P(X = 1) + {v}*P(X = 2)}\\& =& \simplify[all,!collectNumbers]{e ^ { -thismany} + {thismany} * e ^ { -thismany} + {v} * (({thismany} ^ 2 * e ^ { -thismany}) / 2)} \\&=& \var{prob2} \end{eqnarray*} \]

to 3 decimal places.

 
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$\var{thismany}$ % of biscuits made by a baker are chocolate chip cookies.

$\var{number1}$ biscuits are selected at random.

 

a)

Assuming a binomial distribution for $X$ , the number of chocolate chip cookies, write down the values of $n$ and $p$.

$X \sim \operatorname{bin}(n,p)$

$n=$  Expected answer:        $p=$  Expected answer:

Find $\operatorname{E}[X]$ the expected number of chocolate chip cookies in our sample:

$\operatorname{E}[X]=$  Expected answer:

Find the standard deviation for the number of chocolate chip cookies in our sample:

Standard deviation = Expected answer: (to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Find the probability that our selection contains exactly $\var{thisnumber}$ chocolate chip cookies.

$\operatorname{P}(X=\var{thisnumber})=$  Expected answer: (to 3 decimal places).

 

Find the probability that our selection contains no more than 1 chocolate chip cookies.

Probability = Expected answer: (to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

a)

1. $X \sim \operatorname{bin}(\var{number1},\var{prob})$, so $n= \var{number1},\;\;p=\var{prob}$.

2. The expectation is given by $\operatorname{E}[X]=n\times p=\var{number1}\times \var{prob}=\var{number1*prob}$

3. $\operatorname{stdev}(X)=\sqrt{n\times p \times (1-p)}=\sqrt{\var{number1}\times \var{prob} \times \var{1-prob}}=\var{sd}$ to 3 decimal places.

b)

1. \[ \begin{eqnarray*}\operatorname{P}(X = \var{thisnumber}) &=& \dbinom{\var{number1}}{\var{thisnumber}}\times\var{prob}^{\var{thisnumber}}\times(1-\var{prob})^{\var{number1-thisnumber}}\\& =& \var{comb(number1,thisnumber)} \times\var{prob}^{\var{thisnumber}}\times\var{1-prob}^{\var{number1-thisnumber}}\\&=&\var{prob1}\end{eqnarray*} \] to 3 decimal places.

2. 

\[ \begin{eqnarray*}\operatorname{P}(X \leq \var{thatnumber})& =& \simplify[all,!collectNumbers]{P(X = 0) + P(X = 1) + {v}*P(X = 2)}\\& =& \simplify[zeroFactor,zeroTerm,unitFactor]{{1 -prob} ^ {number1}+ {number1} *{prob} *{1 -prob} ^ {number1 -1} + {v} * ({number1} * {number1 -1}/2)* {prob} ^ 2 *( {1 -prob} ^ {number1 -2})}\\& =& \var{prob2}\end{eqnarray*} \]

to 3 decimal places.

 
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The electricity consumption, $X$, of a frozen foods warehouse each week in the summer months  is normally distributed with mean 1250k and standard deviation 90k Wh.

i.e.   \[X \sim \operatorname{N}(\var{m},\var{s}^2)\]

 

Find the probability that in a particular week the electricity consumption is less than 1185 k Wh:

Probability = Expected answer:  (to 4  decimal places)

Find the probability that in a particular week the electricity consumption is greater than 1340 k Wh:

Probability = Expected answer:  (to 4  decimal places)

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

1. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X < {lower}) = P(Z < ({lower} -{m}) / {s}) = P(Z < {lower-m}/{s}) = 1 -P(Z < {m-lower}/{s})} = 1 -\var{p} = \var{precround(1 -p,4)}$ to 4 decimal places.

2. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X > {upper}) = P(Z > ({upper} -{m}) / {s}) = P(Z > {upper-m}/{s}) = 1 -P(Z < {upper-m}/{s})} = 1-\var{p1} = \var{precround(1 -p1,4)}$ to 4 decimal places.
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A new supermarket plans to open somewhere on the outskirts of a town. In fact, $X$, the distance of a new supermarket from the town centre is Uniformly distributed between $\var{lower}$ metres and $\var{upper}$ metres i.e. 

\[X \sim \operatorname{U}(\var{lower},\var{upper})\]

a)

Find $\operatorname{E}[X]$, the expected distance in metres of the new supermarket from the town centre:

$\operatorname{E}[X]=$  Expected answer: m (to 3 decimal places).

Also find the variance $\operatorname{Var}(X)$:

 $\operatorname{Var}(X)=$  Expected answer: (to 3 decimal places).

 

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Find the probability that the supermarket opens within $\var{thisdis}$ kilometres of the town centre.

$P(X \le \var{thisdis}\textrm{km})=$  Expected answer:  (to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

a)

For a uniform distribution \[X \sim \operatorname{U}(\var{lower},\var{upper})\] we have:

$\displaystyle \operatorname{E}[X] = \frac{\var{lower}+\var{upper}}{2}=\var{ans1}$m

$\displaystyle \operatorname{Var}[X] = \frac{(\var{upper}-\var{lower})^2}{12}=\frac{(\var{upper-lower})^2}{12}=\var{ans2}$ to 3 decimal places.

b)

$\displaystyle P(X \le \var{thisdis}\textrm{km})=\frac{\var{thisdis}\times 1000 -\var{lower}}{\var{upper}-\var{lower}}=\var{ans3}$ to 3 decimal places.
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The time,  in minutes between customer arrivals at the RyanJet check-in desk at Newcastle Airport follows an exponential distribution with rate $\var{ra}$ i.e. 

\[X \sim \operatorname{exp}(\var{ra})\]

a)

Find $\operatorname{E}[X]$ between customer arrivals at the RyanJet check-in desk at Newcastle Airport :

$\operatorname{E}[X]=$  Expected answer: minutes (enter as a decimal correct to 3 decimal places).

Find $\operatorname{Var}(X)$:

$\operatorname{Var}(X)=$  Expected answer:  (enter as a decimal correct to 3 decimal places).

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

b)

Find the probability that the time between two customers arriving is less than $\var{thistime}$ minutes:

 $P(X \lt \var{thistime})=$  Expected answer: (enter as a decimal correct to 3 decimal places)

This feedback is on your last submitted answer. Submit your changed answer to get updated feedback.

Advice

If $X \sim \operatorname{exp}(\lambda)$ then $\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}$ and  $\displaystyle \operatorname{Var}(X)=\frac{1}{\lambda^2}$.

Also $P(X \lt a)=1-e^{-\lambda a}$.

a)

If $X \sim \operatorname{exp}(\var{ra})$ then:

$\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}=\frac{1}{\var{ra}}=\var{ans1}$ to 3 decimal places.

$\displaystyle \operatorname{Var}(X) =\frac{1}{\lambda^2}=\frac{1}{\var{ra}^2}=\var{ans2}$ to 3 decimal places.

b)

$P(X \lt \var{thistime}) = 1 -(e ^ {-\var{ ra} \times \var{thistime}}) = 1 -(e ^ { -\var{ra * thistime}}) = \var{ans3}$ to 3 decimal places.
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