Problem 43 - Basic Statistics for Business and Economics

Question

Best Electronics, Inc. offers a "no hassle" returns policy. The number of items returned per day follows the normal distribution. The mean number of customer returns is 10.3 per day and the standard deviation is 2.25 per day. a. In what percent of the days are there 8 or fewer customers returning items? b. In what percent of the days are between 12 and 14 customers returning items? c. Is there any chance of a day with no returns?

Step-by-Step Solution

Answer

a) 15.3%, b) 17.6%, c) Virtually 0% chance.

Step 1: Understanding the Normal Distribution

The returns form a normal distribution with mean \( \mu = 10.3 \) and standard deviation \( \sigma = 2.25 \). We will calculate the probability of different number of returns using the Z-score, which is defined as \( Z = \frac{X - \mu}{\sigma} \).

Step 2: Calculating Percent for 8 or Fewer Returns

For 8 or fewer returns, we need \( P(X \leq 8) \). First, calculate the Z-score for \( X = 8 \): \[ Z = \frac{8 - 10.3}{2.25} \approx -1.022 \]Using the standard normal distribution table, the probability \( P(Z \leq -1.022) \approx 0.153 \), or 15.3%.

Step 3: Calculating Percent for 12 to 14 Returns

For 12 to 14 returns, calculate the Z-scores for both values:- For 12: \[ Z = \frac{12 - 10.3}{2.25} \approx 0.756 \]- For 14: \[ Z = \frac{14 - 10.3}{2.25} \approx 1.644 \]Using the standard normal distribution table, find both probabilities:- \( P(Z \leq 0.756) \approx 0.774 \)- \( P(Z \leq 1.644) \approx 0.950 \)The probability of having between 12 and 14 returns is \( P(0.756 < Z < 1.644) = 0.950 - 0.774 = 0.176 \), or 17.6%.

Step 4: Evaluating the Probability for No Returns

Calculate the Z-score for 0 returns: \[ Z = \frac{0 - 10.3}{2.25} \approx -4.578 \]Since \( Z = -4.578 \) is far in the tail of the normal distribution, the probability \( P(X = 0) \approx 0 \). Thus, there's virtually no chance of a day with no returns.

Key Concepts

Z-score calculations

The Z-score is a fundamental concept used to understand how far a particular data point is from the mean of a dataset expressed in terms of standard deviations. It is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \]Where:

• \(X\) is the value for which you are calculating the Z-score.
• \(\mu\) is the mean of the dataset.
• \(\sigma\) is the standard deviation.

A positive Z-score means the data point is above the mean, while a negative Z-score indicates it's below the mean. For example, in the given situation of Best Electronics, if we want to find the Z-score for 8 returns, we subtract the mean (10.3) from 8 and divide by the standard deviation (2.25). This gives us approximately \(-1.022\), suggesting that 8 is more than one standard deviation below the mean.

probability

Probability is the measure of the likelihood that a certain event will occur. When dealing with a normal distribution, probability is calculated using Z-scores that correspond to areas under the standard normal distribution curve. Each Z-score is linked to a probability that helps us understand the chance of observing a data point within a specific range. For example, in the Best Electronics case, we calculated the probability of receiving 8 or fewer returns. By finding the Z-score for 8 and using a standard normal distribution table, it was determined that there is a 15.3% chance a given day will see 8 or fewer returns. Additionally, the probability of having between 12 and 14 returns was found by calculating the Z-scores for both 12 and 14, then subtracting the smaller cumulative probability from the larger one, resulting in a 17.6% chance.

standard normal distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is a key reference for calculating probabilities in statistics. The standard normal curve, also known as the bell curve, allows us to translate our raw scores (via Z-scores) into probabilities of occurrences. The significance of this distribution is that it enables the conversion of any normal distribution into the standard form through the use of Z-scores. For Best Electronics, returns were initially given in their raw form, and through standardized Z-scores, we could easily find the probabilities associated with different customer return scenarios, such as 8 or fewer returns, or between 12 to 14 returns.

mean and standard deviation

Mean and standard deviation are integral components for understanding normal distributions. The mean, represented as \(\mu\), is the average of all data points. In the Best Electronics case, it is 10.3, indicating that on average, about 10.3 items are returned each day. The standard deviation, represented as \(\sigma\), is a measure of the dispersion of data points around the mean. A smaller standard deviation signals that the data points are closer to the mean, while a larger one indicates more spread. Here, the standard deviation is 2.25, telling us how much the daily returns can vary from the average. Together, they provide a complete picture of the distribution of values and are essential in calculating Z-scores, which in turn are used to calculate probabilities.