Problem 196 - Applied Statistics and Probability for Engineers

Question

The sick-leave time of employees in a firm in a month is normally distributed with a mean of 100 hours and a standard deviation of 20 hours. (a) What is the probability that the sick-leave time for next month will be between 50 and 80 hours? (b) How much time should be budgeted for sick leave if the budgeted amount should be exceeded with a probability of only \(10 \% ?\)

Step-by-Step Solution

Answer

(a) Probability is 0.1525. (b) Budget 125.64 hours for sick leave.

Step 1: Understanding the Normal Distribution

The problem states that the sick-leave time of employees is normally distributed with a mean (\( \mu \)) of 100 hours and a standard deviation (\( \sigma \)) of 20 hours. The task involves calculating probabilities and finding a threshold value (sick-leave time) based on this distribution.

Step 2: Convert Sick-leave Time to Z-scores (Part a)

To find the probability of sick-leave time being between 50 and 80 hours, we convert these values into Z-scores using the formula: \( Z = \frac{X - \mu}{\sigma} \). This yields Z-scores of \( Z_{50} = \frac{50 - 100}{20} = -2.5 \) and \( Z_{80} = \frac{80 - 100}{20} = -1 \).

Step 3: Determine the Probabilities for Z-scores

Consult a standard normal distribution table or use technology to find the probabilities associated with the Z-scores calculated. \( P(Z < -2.5) \approx 0.0062 \) and \( P(Z < -1) \approx 0.1587 \). To find the probability between these two Z-scores, calculate \( P(-2.5 < Z < -1) = 0.1587 - 0.0062 = 0.1525 \).

Step 4: Convert Probability to Sick Leave Time (Part b)

For part (b), we need to find the sick-leave time that corresponds to the 90th percentile of the normal distribution (since only 10% should exceed this time). Using the Z-table, \( Z_{0.90} \approx 1.282 \). Solve for \( X \) in the formula \( X = \mu + Z \cdot \sigma \): \( X = 100 + 1.282 \cdot 20 = 125.64 \) hours.

Key Concepts

Z-scores

Z-scores are a way of describing the position of a raw score in relation to the mean of a distribution. It transforms any value from a normal distribution into a standard normal distribution setting. This is done by converting the raw score into a standardized score, or Z-score. The formula is simple:

• \( Z = \frac{X - \mu}{\sigma} \)

Where:

• \( X \) is the raw score,
• \( \mu \) is the mean of the population,
• and \( \sigma \) is the standard deviation.

By using this formula, any score can be represented as how many standard deviations it lies from the mean. For example, in the exercise provided, a score of 50 hours translates to a Z-score of -2.5 when the mean is 100 hours, and the standard deviation is 20 hours. This means that 50 hours is 2.5 standard deviations below the mean. Understanding Z-scores is critical in using standard normal distribution tables for finding probabilities or percentiles.

Standard Normal Distribution

The standard normal distribution is a specific case of the normal distribution that has a mean of 0 and a standard deviation of 1. It is a way to simplify and standardize calculations that involve normally distributed data. When you convert raw scores into Z-scores, you essentially convert the data into this standard form. Why is this important? Because it allows us to use standardized normal distribution tables or calculators to find probabilities, cumulative areas, or other important statistics. In the problem, once sick-leave times are converted into Z-scores, checking between Z=-2.5 and Z=-1 gives a probability of approximately 0.1525. This value tells us the likelihood of an event happening within specified parameters, like the sick-leave times between 50 and 80 hours. The magic of the standard normal distribution lies in its predictability and simplicity from complex data.

Percentiles in Statistics

Percentiles in statistics give insights into the relative standing of a value in a dataset. They are often used in assessments, like test scores or in this exercise, budget planning.To find what sick-leave time should be planned so that only 10% exceed it, you'd look to find the 90th percentile. This is because 90% of the data is below this point, leaving the upper 10% above it. When dealing with a normal distribution, percentiles can be determined using Z-scores and standard deviation.From the scenario, we know that this equates to a Z-score of roughly 1.282, meaning the value lies 1.282 standard deviations above the mean. Using the transformation formula from Z to raw score, \( X = \mu + Z \cdot \sigma \), it translates to a budgeted sick-leave time of about 125.64 hours. Comprehending percentiles provides a proactive way to manage expectations and allocate resources.