Problem

A delicatessen located in the heart of the business district of a large city serves a vari...

A delicatessen located in the heart of the business district of a large city serves a variety of customers. The delicatessen is open 24 hours a day every day of the week. In an effort to speed up take-out orders, the deli accepts orders by fax. If, on the average, 20 orders are received by fax every two hours throughout the day, find the

a. probability that a faxed order will arrive within the next 9 minutes

b. probability that the time between two faxed orders will be between 3 and 6 minutes

c. probability that 12 or more minutes will elapse between faxed orders

Step-by-Step Solution

Solution 1

The given data pertains to discrete occurrences. Hence, we conclude that the data is exponentially distributed.

Also given that 20 orders are received every two hours. Hence, the mean time between events (in this case, order arrivals) is \(\frac{120}{20}=6\).

That is, \(\frac{1}{\lambda}=6\).

(a)

We need to find the probability that an order will arrive in the next 9 minutes, that is, we need to find \(P(x \leq 9)\).

To find the required probability, we need to use the exponential probability as

\(P(0 \leq x \leq a)=1-e^{-\lambda a}\)

where,

\(a=\) the value of interest

\(\frac{1}{\lambda}=\) mean

\(e=\) natural number

\(=2.71828 .\)

Since we have \(\frac{1}{\lambda}=6\), we get

\(\lambda=0.1667\)

On substituting \(\lambda=0.1667\) and \(a=9\), we get

$$ \begin{aligned} P(0 \leq x \leq 9) &=1-e^{-(0.1667)(9)} \\ &=1-e^{-1.5} . \end{aligned} $$

On using a calculator, we get

$$ \begin{aligned} P(0 \leq x \leq 9) &=1-0.2231 \\ &=0.7769 . \end{aligned} $$

Therefore,

$$ P(x \leq 9)=0.7769 $$

(b)

We need to find \(P(3 \leq x \leq 6)\).

Since \(P(3 \leq x \leq 6)=P(0 \leq x \leq 6)-P(0 \leq x \leq 3)\), first we need to find the probability \(P(0 \leq x \leq 6)\) and \(P(0 \leq x \leq 3)\).

Now, to find these probabilities, we need use the exponential probability as \(P(0 \leq x \leq a)=1-e^{-\lambda a}\)

where

\(a=\) The value of interest.

\(\frac{1}{\lambda}=\) Mean.

\(e=\) Natural number \(=2.71828\).

Since we have \(\frac{1}{\lambda}=6\), we get

\(\lambda=0.1667\)

For \(P(0 \leq x \leq 6)\), we have \(a=6\).

On substituting \(\lambda=0.1667\) and \(a=6\) in \(P(0 \leq x \leq a)=1-e^{-\lambda a}\), we get

$$ \begin{aligned} P(0 \leq x \leq 6) &=1-e^{-(0.1667)(6)} \\ &=1-e^{-1.0002} . \end{aligned} $$

On using a calculator, we get

$$ \begin{aligned} P(0 \leq x \leq 6) &=1-0.3678 \\ &=0.6322 . \end{aligned} $$

For \(P(0 \leq x \leq 3)\), we have \(a=3\).

On substituting \(\lambda=0.1667\) and \(a=3\) in \(P(0 \leq x \leq a)=1-e^{-\lambda a}\), we get

$$ \begin{aligned} P(0 \leq x \leq 3) &=1-e^{-(0.1667)(3)} \\ &=1-e^{-0.5001} . \end{aligned} $$

On using a calculator, we get

\(P(0 \leq x \leq 3)=1-0.6064\)

$$ =0.3936 . $$

Therefore, on substituting \(P(0 \leq x \leq 6)=0.6322\) and \(P(0 \leq x \leq 3)=0.3936\) in

\(\begin{aligned} P(3 \leq x \leq 6) &=P(0 \leq x \leq 6)-P(0 \leq x \leq 3), \text { we get } \\ P(3 \leq x \leq 6) &=0.6322-0.3936 \\ &=0.2386 \end{aligned}\)

(c)

We need to find \(P(x \leq 12)\).

To find the required probability, first we need to find \(P(0 \leq x \leq 12)\) by using the exponential probability as

\(P(0 \leq x \leq a)=1-e^{-\lambda a}\)

where

\(a=\) The value of interest.

\(\frac{1}{\lambda}=\) Mean.

\(e=\) Natural number

\(=2.71828\).

Since we have \(\frac{1}{\lambda}=6\), we get

\(\lambda=0.1667\)

On substituting \(\lambda=0.1667\) and \(a=12\), we get

$$ \begin{aligned} P(0 \leq x \leq 12) &=1-e^{-(0.1667)(12)} \\ &=1-e^{-2.0004} . \end{aligned} $$

On using a calculator, we get

\(P(0 \leq x \leq 12)=1-0.1352\)

$$ =0.8648 $$

Therefore,

$$ \begin{aligned} P(x \geq 12) &=1-P(0 \leq x \leq 12) \\ &=1-0.8648 \\ &=0.1352 . \end{aligned} $$

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