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A distribution of measurements has a mean of 75 and a standard deviation of \(5 .\) You know nothing else about the size or shape of the data. Use this information to find the proportion of measurements in the intervals given. Between 62.5 and 87.5

Short Answer

Expert verified
Answer: Approximately 97.35% of the measurements in the distribution lie between 62.5 and 87.5.

Step by step solution

01

Find the distance from the mean to the interval endpoints in terms of standard deviations

First, we need to find how many standard deviations away from the mean our given interval lies. To do this, subtract the mean from the endpoints of the interval, then divide by the standard deviation: 62.5 - 75 = -12.5 -12.5 / 5 = -2.5 standard deviations 87.5 - 75 = 12.5 12.5 / 5 = 2.5 standard deviations So the interval of 62.5 to 87.5 is -2.5 to 2.5 standard deviations away from the mean.
02

Apply the Empirical Rule

Since we assumed our data follows a normal distribution, we can apply the Empirical Rule to approximate the proportion of measurements in the interval between 62.5 and 87.5. The Empirical Rule states that: - 68% of the data falls within one standard deviation of the mean - 95% of the data falls within two standard deviations of the mean - 99.7% of the data falls within three standard deviations of the mean In our case, the interval lies -2.5 to 2.5 standard deviations away from the mean, which is between the range of two and three standard deviations. As stated above, 95% of the data falls within two standard deviations, and 99.7% within three standard deviations. Since our interval is between these two ranges, we can approximate the proportion of measurements in the interval by interpolating between 95% and 99.7%.
03

Approximate the proportion of measurements in the given interval

To approximate the proportion, we will take the weighted average of 95% and 99.7%, with weights proportional to the distance from the endpoint of our interval to the range of two and three standard deviations. The interval lies 0.5 standard deviations away from the two-standard-deviations range and 0.5 standard deviations away from the three-standard-deviations range. Weighted average: (0.5 * 95% + 0.5 * 99.7%) / (0.5 + 0.5) = (47.5% + 49.85%) / 1 = 97.35% Therefore, approximately 97.35% of the measurements in the distribution lie between 62.5 and 87.5.

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Key Concepts

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

Normal Distribution
A normal distribution, also known as a Gaussian distribution, is a statistical distribution where data points are symmetrically distributed around the mean. Imagine a bell-shaped curve - that's what a normal distribution looks like! Most of the data points tend to cluster around a central point, which is the mean, with fewer points appearing as you move to the tails of the curve. This distribution is hugely important because many natural phenomena tend to follow this pattern.
  • Normal distributions are defined by two key parameters: the mean and the standard deviation.
  • The mean indicates where the center of the data is concentrated.
  • The standard deviation shows how spread out the data points are around the mean.
This concept is essential because it allows us to apply the Empirical Rule, which describes how the data is distributed within a specified number of standard deviations from the mean.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it shows us how much the individual data points differ from the mean of the dataset. The larger the standard deviation, the more spread out the data points are. Here's how to think about it:
  • If you have a low standard deviation, the data points are close to the mean, indicating consistency.
  • With a high standard deviation, the data points are more spread out and there's a higher level of variability.
In our exercise, we have a standard deviation of 5, meaning the average distance between each data point and the mean is 5 units. This is crucial for applying the Empirical Rule and helps us estimate the proportion of data within certain intervals.
Approximate Proportion
When dealing with normal distributions, calculating the exact proportion of measurements within an interval often requires complex statistics. Luckily, the Empirical Rule helps us to approximate these proportions easily. Here's the beauty of this rule in simple steps:
  • About 68% of data will fall within one standard deviation from the mean.
  • About 95% of data will be within two standard deviations.
  • Approximately 99.7% lies within three standard deviations.
In our specific problem, we calculated that the interval from 62.5 to 87.5 encompasses -2.5 to 2.5 standard deviations from the mean. We approximated the proportion of measurements in this range to be roughly 97.35% by interpolating between the percentages given for two and three standard deviations.
Mean
The mean, often referred to as the average, is a central concept in statistics. It is the sum of all the values in a dataset divided by the number of values. The mean tells us where the center of a dataset lies, providing a quick snapshot of where most data points are likely to cluster. To understand it better:
  • It gives us a single representative value for a dataset.
  • The mean is sensitive to extreme values or outliers, which can skew the data.
In our text problem, the mean is 75, indicating that the central value around which our data points are distributed is 75. Understanding the mean helps us relate the calculations of standard deviation and the application of the Empirical Rule.

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Most popular questions from this chapter

For the data sets in Exercises \(1-3,\) calculate the sample variance, \(s^{2}\), using (1) the definition formula and (2) the computing formula. Then calculate the sample standard deviation, s. \(n=5\) measurements: 2,1,1,3,5

Calculate the five-number summary and the interquartile range. Use this information to construct a box plot and identify any outliers. \(n=8\) measurements: .23, .30, .35, .41, .56, .58, .76, .80

For the data sets find the mean, the median, and the mode. Comment on the skewness or symmetry of the data. In a psychology experiment, 10 subjects were given 5 minutes to complete a task. Their time on task (in seconds) is recorded. \(\begin{array}{lllll}175 & 190 & 250 & 230 & 240 \\ 200 & 185 & 190 & 225 & 265\end{array}\)

Find the five-number summary and the IQR. Use this information to construct a box plot and identify any outliers. A count of the number of bacteria (per 100 milliliters) in 10 samples of seawater gave these readings: $$ \begin{array}{llllllllll} 49, & 70, & 54, & 67, & 59, & 40, & 61, & 69, & 71, & 52 \end{array} $$

Ten of the 50 largest businesses in the United States, randomly selected from the Fortune \(500,\) are listed as follows along with their revenues (in millions of dollars) \(^{4}\) : $$\begin{array}{lr|lr}\hline \text { Company } & \text { Revenues } & \text { Company } & \text { Revenues } \\\\\hline \text {GeneralMotors } & \$ 166,380 & \text { Target } & \$ 69,495 \\\\\text { IBM } & 79,919 & \text { Morgan Stanley } & 37,949 \\\\\text { Bank of America } & 93,662 & \text { Johnson \& Johnson } & 71,890 \\\\\text { Home Depot } & 94,595 & \text { Apple } & 215,639 \\\\\text { Boeing } & 94,571 & \text { Exxon Mobil } & 205,004 \\\\\hline\end{array}$$ a. Draw a stem and leaf plot for the data. Are the data skewed? b. Calculate the mean revenue for these 10 businesses. Calculate the median revenue. c. Which of the two measures in part b best describes the center of the data? Explain.

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