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Are the Data Normal? A cidity of Rainfall. Exercise \(3.31\) (page 93 ) concerns the acidity (measured by \(\mathrm{pH}\) ) of rainfall. A sample of 105 rainwater specimens had mean \(\mathrm{pH} 5.43\), standard deviation \(0.54\), and five-number summary \(4.33,5.05,5.44,5.79,6.81 .14\) a. Compare the mean and median and also the distances of the two quartiles from the median. Does it appear that the distribution is quite symmetric? Why? b. If the distribution is really \(N(5.43,0.54)\), what proportion of observations would be less than \(5.05\) ? Less than \(5.79\) ? Do these proportions suggest that the distribution is close to Normal? Why?

Short Answer

Expert verified
The distribution seems quite symmetric as mean and median are close, and quartile distances are similar. The proportions match normality, suggesting a close approximation.

Step by step solution

01

Compare the Mean and Median

To determine symmetry, compare the mean and the median of the dataset. Given the mean is 5.43 and the median (from the five-number summary) is 5.44, notice that the mean and median are almost identical, suggesting potential symmetry.
02

Analyze Quartiles Around the Median

Evaluate the distances from the median to the first quartile (Q1) and to the third quartile (Q3). Q1 = 5.05 (distance from median = 5.44 - 5.05 = 0.39) and Q3 = 5.79 (distance from median = 5.79 - 5.44 = 0.35). Since these distances are quite similar, they suggest that the distribution is reasonably symmetric.
03

Calculate Proportion for Less Than 5.05

Assume a normal distribution with mean 5.43 and standard deviation 0.54. Calculate the z-score for 5.05 using the formula \( z = \frac{X - \mu}{\sigma} \), where \( X = 5.05 \), \( \mu = 5.43 \), and \( \sigma = 0.54 \). The z-score becomes \( z = \frac{5.05 - 5.43}{0.54} \approx -0.70 \). Using a standard normal distribution table, find that \( P(Z < -0.70) \approx 0.2420 \).
04

Calculate Proportion for Less Than 5.79

Similarly, compute the z-score for 5.79: \( z = \frac{5.79 - 5.43}{0.54} \approx 0.67 \). Use the standard normal distribution table to find \( P(Z < 0.67) \approx 0.7486 \). Thus, approximately 74.86% of values fall below 5.79.
05

Evaluate Normality Assumption

Compare the calculated proportions for less than 5.05 and 5.79 to the data quartiles. The quartiles suggest about 25% of data falls below 5.05 and 75% below 5.79. These match closely with the proportions from the normal distribution, supporting the assumption of approximate normality.

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Key Concepts

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

Data Symmetry
When evaluating data for symmetry, one crucial aspect is comparing the mean and median. The mean is the average of all data points, while the median is the middle value when the numbers are arranged in order. For symmetrical data, these two are often close.

In our exercise, the mean is 5.43 and the median, derived from the five-number summary, is 5.44. These values are so close that it suggests the data distribution is symmetrical. Additionally, analyzing the distances of the quartiles from the median provides further insight. If these distances are nearly identical, the distribution is likely symmetrical.

In this case, the distance from the first quartile (5.05) to the median is almost the same as the distance from the third quartile (5.79) to the median. This balance supports the idea that the data is symmetric.
Five-Number Summary
The five-number summary is a simple yet powerful tool to get a quick view of a dataset's distribution. It consists of five key points: the minimum, first quartile (Q1), median, third quartile (Q3), and maximum. This summary is essential in statistics as it provides a concise overview of the data.
  • Minimum: the smallest data value.
  • Q1: the median of the lower half excluding the median of the dataset.
  • Median: the central value of the dataset.
  • Q3: the median of the upper half excluding the median of the dataset.
  • Maximum: the largest data value.
For the rainwater pH data, the five-number summary is given as 4.33, 5.05, 5.44, 5.79, and 6.81 respectively. These values provide insights into the spread and center of the data. They help identify the data range, the center, and any potential outliers or asymmetries.
Z-Score
Z-scores are used to determine how far away a data point is from the mean, in terms of the standard deviation. It tells you where a value stands relative to the average of the dataset. The formula to calculate a z-score is:\[z = \frac{X - \mu}{\sigma}\]where:
  • \( X \) is the actual value you want to find the relative position for,
  • \( \mu \) is the mean of the dataset, and
  • \( \sigma \) is the standard deviation.
In our problem, the z-score for a pH of 5.05 was calculated to be approximately -0.70. This means that 5.05 is 0.70 standard deviations below the mean of 5.43. Similarly, a pH of 5.79 has a z-score of 0.67, indicating that it is around 0.67 standard deviations above the mean. Using these z-scores with a normal distribution table, we can determine the proportion of data points falling below these values. This helps in assessing the normality of the distribution.
Quartiles
Quartiles are numerical values that divide a data set into four equal parts. Each quartile represents a portion of the data relative to its position. The first quartile (Q1) marks the 25th percentile, the second quartile (median) marks the 50th percentile, and the third quartile (Q3) marks the 75th percentile.

These quartiles help in understanding the spread of the data. In contexts like the one from our exercise on rainfall acidity, the differences between quartiles and the median offer insights into symmetry and data distribution behavior. The close distances from Q1 to the median and from Q3 to the median suggest that the data might be symmetrically distributed about the median.

With Q1 at 5.05 and Q3 at 5.79, the data between these quartiles occupies the middle 50% of observations, which is crucial for identifying how clustered or spread out the data is around the median.

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Most popular questions from this chapter

Daily Activity. It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. \(\frac{10}{\text { Among mildly }}\) obese people, minutes of activity varied according to the \(N(373,67)\) distribution. Minutes of activity for lean people had the \(N(526,107)\) distribution. Within what limits do the active minutes for about \(95 \%\) of the people in each group fall? Use the 68-95-99.7 rule.

Monsoon Rains. The summer monsoon rains in India follow approximately a Normal distribution with mean 852 millimeters \((\mathrm{mm}\) ) of rainfall and standard deviation \(82 \mathrm{~mm}\). a. In the drought year \(1987,697 \mathrm{~mm}\) of rain fell. In what percentage of all years will India have \(697 \mathrm{~mm}\) or less of monsoon rain? b. "Normal rainfall" means within \(20 \%\) of the long-term average, or between \(682 \mathrm{~mm}\) and \(1022 \mathrm{~mm}\). In what percentage of all years is the rainfall normal?

Where Are the Quartiles? How many standard deviations above and below the mean do the quartiles of any Normal distribution lie? (Use the standard Normal distribution to answer this question.)

Understanding Density Curves. Remember that it is areas under a density curve, not the height of the curve, that give proportions in a distribution. To illustrate this, sketch a density curve that has a tall, thin peak at 0 on the horizontal axis but has most of its area close to 1 on the horizontal axis without a high peak at 1 .

Upper Arm Lengths. The upper arm length of males over 20 years old in the United States is approximately Normal with mean \(39.1\) centimeters \((\mathrm{cm})\) and standard deviation \(5.0 \mathrm{~cm}\). Use the 68-95-99.7 rule to answer the following questions. (Start by making a sketch like Figure 3.10.) a. What range of lengths covers the middle \(99.7 \%\) of this distribution? b. What percentage of men over 20 have upper arm lengths greater than \(44.1 \mathrm{~cm}\) ?

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