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Chapter 2: Problem 67

Is Michigan Normal? We collected data on the tuition charged by colleges and universities in Michigan. Here are some numerical summaries for the data: $$ \begin{array}{lccc} \text { Mean } & \text { Std. Dev. } & \text { Min } & \text { Max } \\ 10614 & 8049 & 1873 & 30823 \end{array} $$ Based on the relationship between the mean, standard deviation, minimum, and maximum, is it reasonable to believe that the distribution of Michigan tuitions is approximately Normal? Explain.

Short Answer

Expert verified
The data is likely not normally distributed due to significant skewness.

Step by step solution

01

Understanding Normal Distribution

A normal distribution is symmetrically bell-shaped, meaning most data points cluster around the mean, and it tapers outwards equally on both sides. When evaluating if a dataset is approximately normal, we expect that: 1) the mean should be centrally located, 2) there should be symmetry in the lower and upper halves of the data, and 3) the standard deviation should cover the range appropriately.
02

Check Symmetry Centered Around the Mean

Evaluate the symmetry by checking if the mean is centrally positioned between the minimum and maximum values. Calculate the midpoint of the data: \[ \text{Midpoint} = \frac{\text{Min} + \text{Max}}{2} = \frac{1873 + 30823}{2} = 16348 \]Compare this midpoint to the mean of 10614. There's a significant difference, indicating that the mean is not centrally located.
03

Analyze Spread Using Standard Deviation

In a normal distribution, data points within one standard deviation (\(\sigma\)) of the mean account for approximately 68% of the population.Check if the standard deviation adequately covers this distribution: \[ \text{Range covered by one } \sigma = 10614 \pm 8049 = [2565, 18663] \]Compare this interval with the actual data range, [1873, 30823]. The interval is quite different and particularly does not cover the wider range on the higher end.
04

Evaluate Skewness Indicators

In a normal distribution, data ranges tend to be somewhat symmetrically spread from the mean. Look at the possibility of skewness: - The data range [1873, 30823] - skewed by visual inspection. - With mean much closer to Min than Max, this often implies right skewness, and not a normal distribution.

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Key Concepts

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

Understanding Skewness
Skewness is a measure of asymmetry in a distribution. In a perfectly normal distribution, the data should be symmetric around the mean, meaning it looks the same to the left and right. However, when a distribution is skewed, it leans more towards one side.
  • **Right Skewness**: When the mean is greater than the median, it's often a sign of right skewness or positive skew. This means that there are a few higher values pulling the average up.
  • **Left Skewness**: Conversely, if the mean is less than the median, the distribution is left-skewed or negatively skewed, indicating fewer lower values pulling the average down.

The patterns in the Michigan tuition data suggest a right skewness. This is evident because the mean tuition cost is closer to the minimum ($1873$) than the maximum ($30823$). This implies that more tuition values are lower rather than being evenly distributed, dismissing the possibility of a normal distribution.
Exploring Standard Deviation
Standard deviation (\(\sigma\)) is a crucial concept in understanding the spread of data around the mean. It provides insight into data variability and how tightly the data points cluster. A smaller standard deviation indicates that data points are closer to the mean, while a larger standard deviation shows more spread out data.
  • In a normal distribution, about \(68\%\) of the data falls within one standard deviation of the mean.
  • The standard deviation can assist with detecting data that spreads out across a wider range than typical bell curve principles would suggest.

In examining Michigan tuitions, with a mean of \(10614\) and a standard deviation of \(8049\), there's an issue. Calculating the range covered by one standard deviation (\([2565, 18663]\)) shows it does not encompass the actual data range from \(1873\) to \(30823\). This discrepancy hints that the data distribution is not only wider than expected under normal conditions, but also indicates a significant deviation from the classic bell-shaped curve.
Mean as a Measure of Central Tendency
The mean, commonly known as the average, is a key measure of central tendency and represents the center of a data set. It sums all values and divides by their number, offering a single number representation of a dataset's central value. However, in resilience to skewness, the mean might not always reflect the true "center".
  • When examining tuition fees, the mean of \(10614\) is certainly useful for summarizing average costs.
  • Yet, if there's significant skewness, the mean could mislead about true central costs.

For Michigan's colleges, with such a pronounced skew, the mean is closer to a smaller value and fails to sit in the midpoint (\(16348\)). This highlights that while mean is informative, it lacks completeness when depicting datasets with unusual spreads or skews.

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Most popular questions from this chapter

Select the best answer A different species of cockroach has weights that follow a Normal distribution with a mean of 50 grams. After measuring the weights of many of these cockroaches, a lab assistant reports that \(14 \%\) of the cockroaches weigh more than 55 grams. Based on this report, what is the approximate standard deviation of weights for this species of cockroaches? (a) 4.6 (b) 5.0 (c) 6.2 (d) 14.0 (e) Cannot determine without more information.

Mr. Olsen uses an unusual grading system in his class. After each test, he transforms the scores to have a mean of 0 and a standard deviation of \(1 . \mathrm{Mr}\). Olsen then assigns a grade to each student based on the transformed score. On his most recent test, the class's scores had a mean of 68 and a standard deviation of 15\. What transformations should he apply to each test score? Explain.

Run fast Peter is a star runner on the track team. In the league championship meet, Peter records a time that would fall at the 80 th percentile of all his race times that season. But his performance places him at the 50 th percentile in the league championship meet. Explain how this is possible. (Remember that lower times are better in this case!)

Length of pregnancies The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. (a) At what percentile is a pregnancy that lasts 240 days (that's about 8 months)? (b) What percent of pregnancies last between 240 and 270 days (roughly between 8 months and 9 months)? (c) How long do the longest \(20 \%\) of pregnancies last?

Density curves Sketch a density curve that might describe a distribution that has a single peak and is skewed to the left.
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