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Are the data Normal? Monsoon rains. Here are the amounts of summer monsoon rainfall (millimeters) for India in the 100 years 1901 to \(2000:^{16}\) :II MONSOON $$ \begin{array}{llllllrrrr} 722.4 & 792.2 & 861.3 & 750.6 & 716.8 & 885.5 & 777.9 & 897.5 & 889.6 & 935.4 \\\ 736.8 & 806.4 & 784.8 & 898.5 & 781.0 & 951.1 & 1004.7 & 651.2 & 885.0 & 719.4 \\\ 866.2 & 869.4 & 823.5 & 863.0 & 804.0 & 903.1 & 853.5 & 768.2 & 821.5 & 804.9 \\\ 877.6 & 803.8 & 976.2 & 913.8 & 843.9 & 908.7 & 842.4 & 908.6 & 789.9 & 853.6 \\\ 728.7 & 958.1 & 868.6 & 920.8 & 911.3 & 904.0 & 945.9 & 874.3 & 904.2 & 877.3 \end{array} $$ \(\begin{array}{llllllll}739.2 & 793.3 & 923.4 & 885.8 & 930.5 & 983.6 & 789.0 & 889.6\end{array}\) \(\begin{array}{lllllll}1020.5 & 810.0 & 858.1 & 922.8 & 709.6 & 740.2 & 860.3\end{array}\) \(\begin{array}{lllllll}887.0 & 653.1913 .6 & 748.3 & 963.0 & 857.0 & 883.4909 .5 & 708.0\end{array}\) \(\begin{array}{lllllll}852.4735 .6955 .9836 .9 & 760.0 & 743.2 & 697.4961 .7 & 866.9908 .8\end{array}\) \(\begin{array}{lllllll}784.7 & 785.0 & 896.6 & 938.4 & 826.4 & 857.3 & 870.5\end{array}\) (a) Make a histogram of these rainfall amounts. Find the mean and the median. (b) Although the distribution is reasonably Normal, your work shows some departure from Normality. In what way are the data not Normal?

Short Answer

Expert verified
The data is approximately normal but may show some skewness and outliers.

Step by step solution

01

Organize the Data

First, list all the rainfall data to get an overview. Ensure to remove any potential formatting errors during the extraction process.
02

Plot a Histogram

Use the data to construct a histogram. This will be done by dividing the range of rainfall amounts into bins and counting how many data points fall into each bin. This helps in visually assessing the shape of the distribution.
03

Calculate the Mean

Find the mean by adding all the rainfall amounts together and dividing by the total number of observations. If the total of the amounts is represented as \( \Sigma X \) and the number of observations is \( N \), then Mean \( \mu = \frac{\Sigma X}{N} \).
04

Calculate the Median

To find the median, first sort the rainfall amounts in ascending order. The median is the middle value of this ordered list if the number of data points is odd, or the average of the two middle values if even.
05

Analyze the Normality of the Data

Compare the histogram's shape to a Normal distribution (bell curve). Check for symmetry, and the mean should ideally be equal to the median. Alternatively, use statistical tests like the Shapiro-Wilk test for a more quantitative analysis. Determine where the data deviates from normality: skewness, kurtosis, or outliers.

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Key Concepts

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

Histogram Construction
One effective way to understand data distribution is by constructing a histogram. This graphical representation allows you to visualize the frequencies of different ranges of data, also known as bins. In the context of the monsoon rainfall data, you start by organizing the rainfall measurements from the smallest to the largest value. Next, you'll need to determine the number and width of the bins. Typically, the number of bins can be decided based on the audience's preference or using formulas like Sturges' rule, which calculates the optimal number based on the number of observations. Once the bins are determined, plot the histogram:
  • Each bar represents the frequency of data points within the bin.
  • The height of the bar illustrates how many data points fall within each range of rainfall amounts.
  • Ensure that all data points are accounted for within the range of your histogram.
This histogram will help you see if there's a bell-shaped, uniform, or skewed distribution. A normal distribution would typically form a bell curve.
Mean and Median Calculation
Calculating the mean and median of a dataset gives insight into its central tendency. Let's start with the mean, which is the average value of the data. Add up all the rainfall measurements and divide by the number of years to find the mean. This formula is expressed as \( \mu = \frac{\Sigma X}{N} \), where \( \Sigma X \) is the total millimeters of rainfall, and \( N \) is the 100 years over which measurements were taken. Next, calculate the median to find the middle point in an ordered list of the data. For a sorted list:
  • If the number of observations \( N \) is odd, the median is the middle value.
  • If \( N \) is even, the median is the average of the two central numbers.
While the mean provides a picture of the dataset's overall trend, the median can be more informative for skewed data, offering a better sense of the center.
Normal Distribution Analysis
Detecting normality in a dataset involves comparing its distribution to the ideal bell curve shape. In the case of the rainfall data, begin by analyzing the histogram. A normal distribution is symmetric around the mean, with the mean, median, and mode all roughly equal. However, real-world data rarely fits perfectly into this mold. Look for:
  • Symmetry: A normal distribution will have a histogram that's mirrored on either side of the mean.
  • Tails: Check if the tails of the data (extreme high and low points) taper off smoothly.
  • Peaks: The peak should occur near the mean value.
Additionally, you can use statistical tests like the Shapiro-Wilk test to provide a quantitative measure of normality. Outliers, skewness (when one tail is longer), and kurtosis (tailedness or peakedness) can indicate departures from normality. These characteristics should prompt closer investigation as they could affect further statistical analyses and outcomes.

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Most popular questions from this chapter

Runners. In a study of exercise, a large group of male runners walk on a treadmill for 6 minutes. Their heart rates in beats per minute at the end vary from runner to runner according to the \(N(104,12.5)\) distribution. The heart rates for male nonrunners after the same exercise have the \(N(130,17)\) distribution. (a) What percent of the runners have heart rates above 140 ? (b) What percent of the nonrunners have heart rates above 140 ?

Heights of men and women. The heights of women aged 20-29 follow approximately the \(N(64.2,2.8)\) distribution. Men the same age have heights distributed as \(N(69.4,3.0)\). What percent of men aged \(20-29\) are taller than the mean height of women aged 20-29?

The scores of adults on an IQ test are approximately Normal with mean 100 and standard deviation 15. Alysha scores 135 on such a test. Her z-score is about (a) \(1.33\). (b) \(2.33\) (c) 6.33.

Normal is only approximate: ACT scores. Composite scores on the ACT test for the 2015 high school graduating class had mean \(21.0\) and standard deviation 5.5. In all, 1,924,436 students in this class took the test. Of these, 205,584 had scores higher than 28 , and another 60,551 had scores exactly 28 . ACT scores are always whole numbers. The exactly Normal \(N(21.0,5.5)\) distribution can include any value, not just whole numbers. What is more, there is no area exactly above 28 under the smooth Normal curve. So ACT scores can be only approximately Normal. To illustrate this fact, find (a) the percent of 2015 ACT scores greater than 28 using the actual counts reported. (b) the percent of \(2015 \mathrm{ACT}\) scores greater than or equal to 28 , using the actual counts reported. (c) the percent of observations that are greater than 28 using the \(N(21.0,5.5)\) distribution. (The percent greater than or equal to 28 is the same, because there is no area exactly over 28.)

Are we getting smarter? When the Stanford-Binet IQ test came into use in 1932 , it was adjusted so that scores for each age group of children followed roughly the Normal distribution with mean 100 and standard deviation 15 . The test is readjusted from time to time to keep the mean at 100 . If present-day American children took the 1932 Stanford-Binet test, their mean score would be about 120 . The reasons for the increase in IQ over time are not known but probably include better childhood nutrition and more experience in taking tests. 11 (a) IQ scores above 130 are often called "very superior." What percentage of children had very superior scores in 1932 ? (b) If present-day children took the 1932 test, what percentage would have very superior scores? (Assume that the standard deviation 15 does not change.)

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