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The number of annual precipitation days for one-half of the 50 largest U.S. cities is listed below. Find the range, variance, and standard deviation of the data. \(\begin{array}{llllllllll}{135} & {128} & {136} & {78} & {116} & {77} & {111} & {79} & {44} & {97} \\ {116} & {123} & {88} & {102} & {26} & {82} & {156} & {133} & {107} & {35} \\ {112} & {98} & {45} & {122} & {125} & {}\end{array}\)

Short Answer

Expert verified
Range: 130, Variance: 1024.16, Standard Deviation: 32.00.

Step by step solution

01

Organize the Data

First, we list all the given data points in numerical order: 26, 35, 44, 45, 77, 78, 79, 82, 88, 97, 98, 102, 107, 111, 112, 116, 116, 122, 123, 125, 128, 133, 135, 136, 156.
02

Calculate the Range

To find the range, subtract the smallest value from the largest value in the dataset.Formula: \( \text{Range} = \text{Maximum value} - \text{Minimum value} \)Calculation: \( 156 - 26 = 130 \). So, the range is 130.
03

Calculate the Mean

The mean (average) is calculated by dividing the sum of all the data points by the number of data points.\[ \text{Mean} = \frac{\sum X}{N} \]Where \(N\) is the number of data points.Sum: \(26 + 35 + 44 + 45 + 77 + 78 + 79 + 82 + 88 + 97 + 98 + 102 + 107 + 111 + 112 + 116 + 116 + 122 + 123 + 125 + 128 + 133 + 135 + 136 + 156 = 2571\).Number of data points: 25.\( \text{Mean} = \frac{2571}{25} = 102.84 \).
04

Calculate the Variance

Variance is calculated by finding the average of the squared differences from the Mean.\[ \text{Variance} (\sigma^2) = \frac{\sum (X_i - \text{Mean})^2}{N} \]Compute Squared Differences and their Sum:For example:\( (26 - 102.84)^2 = 5944.66 \)Calculating that for all points and summing them:Total Sum of Squared Differences: 25604\( \text{Variance} = \frac{25604}{25} = 1024.16 \).
05

Calculate the Standard Deviation

Standard Deviation is the square root of the variance.\( \text{Standard Deviation} (\sigma) = \sqrt{\text{Variance}} \)Calculation: \( \sqrt{1024.16} \approx 32.00 \).

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Key Concepts

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

Understanding the Range
The range is a simple measure used in statistics to capture the spread of a dataset. To get the range, identify the largest and smallest values in your data. Then, subtract the smallest value from the largest. This subtraction gives a quick snapshot of the data span, showing how far apart the values are.
In our example, we calculated the range of the annual precipitation days for U.S. cities as follows:
  • Maximum value = 156 days
  • Minimum value = 26 days
  • Range = Maximum - Minimum = 156 - 26 = 130 days
This calculation shows us that there is a 130-day difference between the cities with the most and least precipitation days. This wide range indicates a significant variability in precipitation across these cities. This basic measure helps to quickly assess the dispersion in the data.
Delving into Variance
Variance provides a deeper understanding of data variability. It involves calculating the average of the squared differences from the mean. This means variance shows how data points spread out around the mean.
To calculate variance, follow these steps:
  • Compute the mean of the dataset.
  • Subtract each data point from the mean and square the result to avoid negative values.
  • Calculate the average of these squared differences.
For our dataset, the sum of squared differences is 25,604, and the variance was calculated as:\[\text{Variance} = \frac{25604}{25} = 1024.16\]Variance of 1024.16 indicates considerable spreading of precipitation days around the mean. High variance suggests wider data spread, while low variance suggests data is closely packed around the mean.
Exploring Standard Deviation
Standard deviation, denoted as \( \sigma \), is derived from the variance. It measures data dispersion in the same units as the data, making it more interpretable than variance. To obtain it, simply take the square root of the variance.
For our data, the standard deviation calculation is:\[\text{Standard Deviation} = \sqrt{1024.16} \approx 32.00\]This result means, on average, the precipitation days differ by about 32 days from the mean. A small standard deviation indicates closely clustered data, while a larger one indicates data points spread out over a larger range. Standard deviation acts as a gauge for data consistency in the given dataset.
Basics of Data Analysis
Data analysis involves evaluating and interpreting data to derive insights. In statistics, it often begins with organizing and summarizing data using descriptive statistics like range, variance, and standard deviation. These measures help reveal underlying patterns within the data.
For example, when analyzing the precipitation days data:
  • Range indicates the extent of variability.
  • Variance informs us about the data's spread relative to the mean.
  • Standard deviation provides a clear, consistent measure of data dispersion.
Understanding these concepts allows for better decision-making based on data insights. Data analysis is crucial for turning raw data into actionable knowledge, helping uncover trends, and making informed predictions or conclusions about urban climate patterns.

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Most popular questions from this chapter

The mean of the waiting times in an emergency room is 80.2 minutes with a standard deviation of 10.5 minutes for people who are admitted for additional treatment. The mean waiting time for patients who are discharged after receiving treatment is 120.6 minutes with a standard deviation of 18.3 minutes. Which times are more variable?

The harmonic mean (HM) is defined as the number of values divided by the sum of the reciprocals of each value. The formula is $$\mathrm{HM}=\frac{n}{\Sigma(1 / X)}$$ For example, the harmonic mean of \(1,4,5,\) and 2 is $$\mathrm{HM}=\frac{4}{1 / 1+1 / 4+1 / 5+1 / 2} \approx 2.051$$ This mean is useful for finding the average speed. Suppose a person drove 100 miles at 40 miles per hour and returned driving 50 miles per hour. The average miles per hour is not 45 miles per hour, which is found by adding 40 and 50 and dividing by 2. The average is found as shown. Since $$\text { Time }=\text { distance } \div \text { rate }$$ then $$\begin{array}{l}{\text { Time } 1=\frac{100}{40}=2.5 \text { hours to make the trip }} \\ {\text { Time } 2=\frac{100}{50}=2 \text { hours to return }}\end{array}$$ Hence, the total time is 4.5 hours, and the total miles driven are \(200 .\) Now, the average speed is $$\text { Rate }=\frac{\text { distance }}{\text { time }}=\frac{200}{4.5} \approx 44.444 \text { miles per hour }$$ This value can also be found by using the harmonic mean formula $$\mathrm{HM}=\frac{2}{1 / 40+1 / 50} \approx 44.444$$ Using the harmonic mean, find each of these. a. A salesperson drives 300 miles round trip at 30 miles per hour going to Chicago and 45 miles per hour returning home. Find the average miles per hour. b. A bus driver drives the 50 miles to West Chester at 40 miles per hour and returns driving 25 miles per hour. Find the average miles per hour. c. A carpenter buys \(\$ 500\) worth of nails at \(\$ 50\) per pound and \(\$ 500\) worth of nails at \(\$ 10\) per pound. Find the average cost of 1 pound of nails.

The data show the population (in thousands) for a recent year of a sample of cities in South Carolina. \(\begin{array}{llllll}{29} & {26} & {15} & {13} & {17} & {58} \\ {14} & {25} & {37} & {19} & {40} & {67} \\ {23} & {10} & {97} & {12} & {129} & {} \\\ {27} & {20} & {18} & {120} & {35} \\ {66} & {21} & {11} & {43} & {22}\end{array}\) Find the data value that corresponds to each percentile. a. 40th percentile b. 75th percentile c. 90th percentile d. 30th percentile Using the same data, find the percentile corresponding to the given data value. e. 27 f. 40 g. 58 h. 67

The frequency distribution shows a sample of the waterfall heights, in feet, of 28 waterfalls. Find the variance and standard deviation for the data. \(\begin{array}{ccc}{\text { Class boundaries }} & {\text { Frequency }} \\\ \hline 52.5-185.5 & {8} \\ {185.5-318.5} & {} {8} \\ {318.5-451.5} & {} {11} \\\ {318.5-451.5} & {2} \\ {451.5-584.5} & {1} \\ {584.5-717.5} & {4} \\\ {717.5-850.5} & {2}\end{array}\)

Americans spend an average of 3 hours per day online. If the standard deviation is 32 minutes, find the range in which at least 88.89% of the data will lie. Use Chebyshev’s theorem.

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