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The average full time faculty member in a post secondary degree-granting institution works an average of 53 hours per week. a. If we assume the standard deviation is 2.8 hours, what percentage of faculty members work more than 58.6 hours a week? b. If we assume a bell-shaped distribution, what percentage of faculty members work more than 58.6 hours a week?

Short Answer

Expert verified
About 2.28% of faculty members work more than 58.6 hours a week.

Step by step solution

01

Understanding the Problem

We are given the average work hours of faculty members as 53 per week with a standard deviation of 2.8 hours. We need to find the percentage of faculty members working more than 58.6 hours, assuming a normal distribution.
02

Convert to Z-Score

First, we convert 58.6 hours into a z-score using the formula \( z = \frac{x - \mu}{\sigma} \), where \( x = 58.6 \), \( \mu = 53 \), and \( \sigma = 2.8 \).
03

Calculate Z-Score

Plug in the values into the formula: \( z = \frac{58.6 - 53}{2.8} = 2 \). So, a work week of 58.6 hours is 2 standard deviations above the mean.
04

Find Percentage Using Z-Score Table

Using the Z-Score table, find the probability (percentage) corresponding to a z-score of 2. The z-score table shows that the cumulative probability for z = 2 is approximately 0.9772.
05

Calculate Percentage Greater Than Z

To find the percentage of faculty working more than 58.6 hours, calculate \( 1 - 0.9772 = 0.0228 \), or 2.28%.
06

Interpret Results for Bell-Shaped Distribution

Since the data follows a normal distribution, the previous steps provide the percentage. Therefore, about 2.28% of faculty members work more than 58.6 hours per week, assuming a bell-shaped distribution.

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Key Concepts

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

Z-Score
The Z-Score is a crucial concept in statistics used to determine how far away a point is from the mean. It plays a vital role in normal distribution problems like the one we're discussing. In our context, the Z-Score reveals how many standard deviations a particular measurement (like our 58.6 hours) is from the average (mean) work week of 53 hours.

To calculate a Z-Score, you can use the formula:
  • \( z = \frac{x - \mu}{\sigma} \)
  • Where \( x \) is the value of interest (58.6 hours in our case).
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation of the dataset.
Using these definitions, after plugging in the numbers, we found that the Z-Score for 58.6 hours is 2. This means 58.6 is two standard deviations above the average work week.Understanding Z-Scores can help you compare different data points relative to their average, especially useful in a bell-shaped or normal distribution.
Standard Deviation
Standard Deviation is a measure of how spread out the numbers in a data set are. For example, in our exercise, it's given as 2.8 hours. This tells us that, on average, the work hours of faculty members deviate from the mean by around 2.8 hours. This measure is crucial because it helps define the "normality" of a distribution, contributing to our ability to calculate Z-Scores effectively. A small standard deviation indicates that the data points are close to the mean, while a large one suggests they're more spread out.

When dealing with a bell-shaped curve or normal distribution, this spread determines how quickly the frequency of data points decreases as you move away from the mean. To put it simply, in a normal distribution:
  • Approximately 68% of data points lie within 1 standard deviation of the mean.
  • About 95% are within 2 standard deviations.
  • Almost 99.7% are within 3 standard deviations.
These percentages can help you quickly assess probabilities, as we did in our exercise. Calculating with standard deviation lets you better understand data variability and predict outcomes more accurately.
Cumulative Probability
Cumulative probability is a way to determine the likelihood that a variable will fall within a certain range or be less than a specified value. Seen often in the context of a normal distribution, it lets us calculate the probability that a random variable is less than or equal to a particular value. In our step-by-step solution, once we calculated the Z-Score for 58.6 hours as 2, we turned to a Z-Score table to find its cumulative probability. The table tells us that the cumulative probability corresponding to a Z-Score of 2 is approximately 0.9772.

This number represents the probability that a faculty member works 58.6 hours or less per week. To find those working more, you subtract this cumulative probability from 1 (i.e., 1 - 0.9772 = 0.0228), yielding a probability of 2.28%. Cumulative probability is useful as it helps to determine thresholds, make predictions, and understand the likelihood of events within a data set. In scenarios involving normal distribution, it assists in estimating the percentage of data within certain ranges, broadening your insight into the underlying patterns.

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