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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
Experience has shown that two in every ten components produced by a circuitboard company will be defective. A random sample of 100 components is inspected for defects, and D is the number of defectives in this sample.
One in ten new small businesses in the north-east goes bust within a year. Let X be the number of small businesses that fail in the next year out of thirty that have been set up.
The number of calls, Y, received at the British Passport Office in Durham occurs at the rate of 10 a minute.
Expected answer:
Binomial DistributionPoisson Distribution
Experience has shown that two in every ten components produced by a circuitboard company will be defective. A random sample of 100 components is inspected for defects, and D is the number of defectives in this sample.
One in ten new small businesses in the north-east goes bust within a year. Let X be the number of small businesses that fail in the next year out of thirty that have been set up.
The number of calls, Y, received at the British Passport Office in Durham occurs at the rate of 10 a minute.

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Advice

No solution given.
\( \begingroup \)

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

b)

Remember that for a Poisson random variable:
\begin{align}
\operatorname{P}(X=r)&=\dfrac{\lambda^r\times e^{-\lambda}}{r!}\\
\end{align}

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

 
\( \endgroup \)
\( \begingroup \)

$\var{thismany}$ % of biscuits made by Fredds, the popular bakery chain, 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. $\operatorname{SD}(X)=\sqrt{n\times p \times (1-p)}=\sqrt{\var{number1}\times \var{prob} \times \var{1-prob}}=\var{sd}$ 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.

 
\( \endgroup \)