Problem 46 - Basic Statistics for Business and Economics

Question

The funds dispensed at the ATM machine located near the checkout line at the Kroger's in Union, Kentucky, follows a normal distribution with a mean of \(\$ 4,200\) per day and a standard deviation of \(\$ 720\) per day. The machine is programmed to notify the nearby bank if the amount dispensed is very low (less than \(\$ 2,500)\) or very high (more than \(\$ 6,000)\). a. What percent of the days will the bank be notified because the amount dispensed is very low? b. What percent of the time will the bank be notified because the amount dispensed is high? c. What percent of the time will the bank not be notified regarding the amount of funds dispersed?

Step-by-Step Solution

Answer

The bank is notified 0.91% for low, 0.62% for high; not notified 98.47% of the time.

Step 1: Understand the Problem

We need to find the percentage of days the bank will be notified based on the distribution of funds dispensed, which follows a normal distribution with \( \mu = 4200 \) and \( \sigma = 720 \). Notifications occur for amounts less than \\(2500 or more than \\)6000.

Step 2: Calculate Z-Score for Low Amount

To find the percentage of days with dispensed amounts less than \$2500, calculate the Z-score: \( Z = \frac{X - \mu}{\sigma} \). For \( X = 2500 \): \[ Z = \frac{2500 - 4200}{720} = \frac{-1700}{720} \approx -2.36 \].

Step 3: Find Probability for Low Amount Using Z-Table

Using the Z-table, we find the probability for \( Z = -2.36 \). The table gives us \( P(Z < -2.36) \approx 0.0091 \), meaning approximately 0.91% of days have low dispensed amounts.

Step 4: Calculate Z-Score for High Amount

For amounts greater than \$6000, calculate the Z-score: \[ Z = \frac{6000 - 4200}{720} = \frac{1800}{720} \approx 2.5 \].

Step 5: Find Probability for High Amount Using Z-Table

Using the Z-table, find the probability of \( Z < 2.5 \), which is \( P(Z < 2.5) \approx 0.9938 \). The probability for \( Z > 2.5 \) is \( 1 - 0.9938 = 0.0062 \) or 0.62%.

Step 6: Combine Probabilities for Notifications

Combine the probabilities for very low and very high dispense: \[ 0.91\% + 0.62\% = 1.53\% \]. Thus, bank notifications occur about 1.53% of the time.

Step 7: Find Probability of No Notification

Subtract the notification probability from 100%: \[ 100\% - 1.53\% = 98.47\% \]. The bank is not notified 98.47% of the time.

Key Concepts

Z-score

In statistics, the Z-score is a helpful metric used to determine how far away a specific data point is from the mean of a data set. It is essentially a measure of how many standard deviations a particular value is from the mean.
To calculate the Z-score, you use the formula:

• Z = \( \frac{X - \mu}{\sigma} \)

Here, \( X \) represents the data point you're examining, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.
For example, if we want to find the Z-score for an amount of \(2500, given a mean of \)4200 and a standard deviation of \(720, we perform the following calculation:

• \( Z = \frac{2500 - 4200}{720} \approx -2.36 \)

The negative Z-score indicates that \)2500 is below the mean.
Z-scores are crucial for understanding the context within normal distribution, helping us find probabilities for specific outcomes efficiently.

probability calculation

Probability calculation is the process of calculating the likelihood of different outcomes.
When using the normal distribution along with the Z-score, you often rely on a Z-table, which helps find the probability that a random variable is less than or equal to a particular value.
Let us consider probabilities for very low dispensals here. For a Z-score of \(-2.36\), which is a low value relative to the mean, you would use a Z-table to find the probability.
The Z-table returns \( P(Z < -2.36) \approx 0.0091 \), indicating the likelihood of dispensing less than \(2500 is approximately 0.91%.

• Reading the Z-table allows us to easily interpret the data and convert it into actionable insights.
• Similarly, to find the probability of dispensing over \)6000, with a Z-score of \(2.5\), the Z-table gives us \( P(Z < 2.5) \approx 0.9938 \).
• Thus, the probability of \( Z > 2.5 \) is \( 1 - 0.9938 = 0.0062 \) or 0.62%.

The probability calculations are not just limited to events like low or high dispensals, but are used across many fields to predict outcomes based on different normal distributions.

standard deviation

Standard deviation is a statistical metric that quantifies the amount of variation or dispersion in a set of values.
If the standard deviation is lower, the data points tend to be closer to the mean, indicating less variability. Conversely, a higher standard deviation indicates more spread out data.

• It is calculated using the formula: \( \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i - \mu)^2} \)

Where:

• \( N \) is the number of data points,
• \( X_i \) represents each individual data point,
• \( \mu \) is the mean of the data set.

In the context of our exercise, the standard deviation of \\(720 applies to the amount dispensed daily.
This measure helps determine how frequently data points fall near or far from the average value.
A standard deviation contextualizes the significance of extreme values, helping us understand that days with very low or very high dispensals are relatively uncommon due to their deviation from the mean of \\)4200.