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Customers arrive at the automatic teller machine in accordance with a Poisson process with rate 12 per hour. The amount of money withdrawn on each transaction is a random variable with mean \(\$ 30\) and standard deviation \(\$ 50 .\) (A negative withdrawal means that money was deposited.) The machine is in use for 15 hours daily. Approximate the probability that the total daily withdrawal is less than \(\$ 6000\).

Short Answer

Expert verified
The approximate probability that the total daily withdrawal is less than $\$6000$ is 0.9452 or 94.52%. This is calculated using the Central Limit Theorem, which shows that the distribution of the total daily withdrawal can be approximated as a normal distribution with a mean of $\$5400$ and a standard deviation of $375.97$.

Step by step solution

01

1. Calculate the number of customer arrivals in 15 hours

Since customer arrivals follow a Poisson process with a rate of 12 per hour, we can calculate the number of arrivals during the 15-hour period. Total arrivals = Arrival rate * Time in hours Total arrivals = 12 per hour * 15 hours = 180 arrivals
02

2. Calculate expected daily withdrawal

The expectation for the total daily withdrawal is the sum of the expected withdrawal on a single transaction over all transactions in a day. Expected daily withdrawal = Expected withdrawal on a single transaction * Total arrivals Expected daily withdrawal = $30 * 180 = \$ 5400
03

3. Calculate the standard deviation of total daily withdrawal

Next, we calculate the standard deviation of the total daily withdrawals using the given standard deviation of a single transaction ($50) and the total number of transactions (180). Total daily withdrawal standard deviation = \(\sqrt{Total \; Arrivals}\) * standard deviation of a single transaction Total daily withdrawal standard deviation = \(\sqrt{180} * \) 50 ≈ $375.97
04

4. Apply the Central Limit Theorem

By the Central Limit Theorem (CLT), the distribution of the total daily withdrawal (assuming independence of the transactions) can be approximated as a normal distribution with the mean and standard deviation calculated in steps 2 and 3 (Expected daily withdrawal: \(5400, and Total daily withdrawal standard deviation: \)375.97).
05

5. Calculate the Z-score for \$6000

Now, we calculate the Z-score for the total daily withdrawal being less than $6000. Z-score = \(\frac{x - \mu}{\sigma}\) Z-score = \(\frac{6000 - 5400}{375.97}\) ≈ 1.60
06

6. Estimate the probability using Z-score

Using a Z-table or calculator, we can find the probability that the total daily withdrawal is less than $6000 by looking up the probability corresponding to the calculated Z-score (1.60). Probability = P(Z < 1.60) ≈ 0.9452 So, the approximate probability that the total daily withdrawal is less than $6000 is 0.9452 or 94.52%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem
When studying random processes, such as the arrival of customers at an ATM, the Central Limit Theorem (CLT) is a statistical powerhouse. It tells us that regardless of the original distribution of a dataset (as long as the data has a well-defined variance), the means of sufficiently large samples will follow a normal distribution. This is crucial for approximation because it allows us to apply techniques that are specific to normal distributions even when the underlying data does not fit that pattern. For example, in the ATM scenario, each customer withdrawal is a random variable with its own mean and standard deviation, but thanks to the CLT, we know that the distribution of the total daily withdrawals can be treated as normal if the number of transactions is large enough, which in this case, it is.

The beauty of CLT lies in its simplicity: it requires the transactions to be independent and identically distributed, which suits many real-world situations. Its implications are vast, making otherwise complex problems manageable, and it forms a bedrock for inferential statistical analysis, which is the cornerstone of making predictions and decisions based on data.
Normal Distribution Approximation
Given that the CLT allows us to use the normal distribution as an approximation, understanding the normal distribution approximation is essential. The normal distribution is a continuous probability distribution that is symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In the case of the ATM, where each customer's withdrawal is random and the total number of daily transactions is high, the sum of these random variables will approximate a normal distribution.

It's akin to the shape of a bell curve, and it presents a clear framework for solving complex problems. Since random variables associated with natural phenomena, human behavior, or measurement errors often end up following a normal distribution, approximating a given situation to normal allows us to predict probabilities and outcomes using tables or computational tools. This enables students and statisticians alike to tackle problems like predicting ATM withdrawals with heightened accuracy and ease.
Standard Deviation Calculation
The standard deviation calculation is a fundamental concept in statistics, offering a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. When we calculate the standard deviation for a total figure, such as daily ATM withdrawals, we multiply the standard deviation of a single transaction by the square root of the total number of transactions. This scales up the individual variability to the level of the overall process we're examining.

Importantly, understanding how to calculate the standard deviation is instrumental for gauging risk and volatility - for instance, in our ATM problem, knowing the standard deviation of withdrawals helps us estimate how typically a day's total might deviate from the average, thus providing a sense of the expected variability and aiding in financial planning and analysis.
Z-score Estimation
When we work with normal distributions, the Z-score estimation is a method of describing a data point's relationship to the mean of a group of data points. Expressed in terms of standard deviations from the mean, the Z-score is a powerful tool because it allows us to convert any score in our data to a common metric that can be easily compared across different datasets. To compute a Z-score in our ATM example, we subtract the mean from the value we're interested in, and then divide this difference by the standard deviation.

This score then represents the number of standard deviations away from the mean – a positive score indicates a value above the mean, while a negative score signifies a value below the mean. Understanding Z-scores is crucial because they tell us not just if a value is above or below average, but how unusual or extreme the value is relative to the overall distribution.
Probability Calculation Using Z-table
The probability calculation using a Z-table is a practical application of the Z-score that lets us determine the probability of a value occurring within a normal distribution. A Z-table lists the cumulative probability associated with each Z-score, effectively telling us the likelihood of encountering a value at or below a given Z-score in a standard normal distribution. This data is indispensable when we approximate complex processes, like the total ATM withdrawals, using normal distributions.

In our ATM case, after finding the Z-score for a withdrawal of $6000, we use the Z-table to find the corresponding cumulative probability, which gives us the probability that the total withdrawals will be less than this amount. For students and statisticians, this step is the culmination of their statistical analysis - translating a Z-score into a real-world probability, which allows for informed decision-making based on solid statistical evidence.

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Most popular questions from this chapter

A set of \(n\) cities is to be connected via communication links. The cost to construct a link between cities \(i\) and \(j\) is \(C_{i j}, i \neq j .\) Enough links should be constructed so that for each pair of cities there is a path of links that connects them. As a result, only \(n-1\) links need be constructed. A minimal cost algorithm for solving this problem (known as the minimal spanning tree problem) first constructs the cheapest of all the (in) links. Then, at each additional stage it chooses the cheapest link that connects a city without any links to one with links. That is, if the first link is between cities 1 and 2, then the second link will either be between 1 and one of the links \(3, \ldots, n\) or between 2 and one of the links \(3, \ldots, n .\) Suppose that all of the \(\left(\begin{array}{c}n \\\ 2\end{array}\right)\) costs \(C_{i j}\) are independent exponential random variables with mean \(1 .\) Find the expected cost of the preceding algorithm if (a) \(n=3\), (b) \(n=4\).

Events occur according to a Poisson process with rate \(\lambda .\) Each time an event occurs, we must decide whether or not to stop, with our objective being to stop at the last event to occur prior to some specified time \(T\), where \(T>1 / \lambda\). That is, if an event occurs at time \(t, 0 \leqslant t \leqslant T\), and we decide to stop, then we win if there are no additional events by time \(T\), and we lose otherwise. If we do not stop when an event occurs and no additional events occur by time \(T\), then we lose. Also, if no events occur by time \(T\), then we lose. Consider the strategy that stops at the first event to occur after some fixed time \(s, 0 \leqslant s \leqslant T\). (a) Using this strategy, what is the probability of winning? (b) What value of \(s\) maximizes the probability of winning? (c) Show that one's probability of winning when using the preceding strategy with the value of \(s\) specified in part (b) is \(1 / e\).

Let \(\left\\{M_{i}(t), t \geqslant 0\right\\}, i=1,2,3\) be independent Poisson processes with respective rates \(\lambda_{i}, i=1,2\), and set $$ N_{1}(t)=M_{1}(t)+M_{2}(t), \quad N_{2}(t)=M_{2}(t)+M_{3}(t) $$ The stochastic process \(\left\\{\left(N_{1}(t), N_{2}(t)\right), t \geqslant 0\right\\}\) is called a bivariate Poisson process. (a) Find \(P\left[N_{1}(t)=n, N_{2}(t)=m\right\\}\) (b) Find \(\operatorname{Cov}\left(N_{1}(t), N_{2}(t)\right)\)

Let \(X\) be an exponential random variable. Without any computations, tell which one of the following is correct. Explain your answer. (a) \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\) (b) \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\) (c) \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\)

Customers arrive at a two-server service station according to a Poisson process with rate \(\lambda .\) Whenever a new customer arrives, any customer that is in the system immediately departs. A new arrival enters service first with server 1 and then with server 2\. If the service times at the servers are independent exponentials with respective rates \(\mu_{1}\) and \(\mu_{2}\), what proportion of entering customers completes their service with server 2?

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