Problem 15 - Calculus Concepts

Question

A quick approximation is sometimes useful when an exact answer is not required. a. Verify the statements in the Empirical Rule for the normal probability density function with \(\mu=5.3\) and \(\sigma=8.372\) b. Use the Empirical Rule, to estimate $$ P(-11.444

Step-by-Step Solution

Answer

a: Verified using intervals; b: Estimate ~81.5%; c: Exact is 81.85%.

Step 1: Understand the Empirical Rule

The Empirical Rule states for a normal distribution, approximately 68% of the data falls within one standard deviation (\(\sigma\)) of the mean (\(\mu\)), about 95% within two standard deviations, and nearly 99.7% within three standard deviations.

Step 2: Calculate Ranges for the Rule Verification

To verify the Empirical Rule, calculate the intervals:- One standard deviation: \((\mu - \sigma, \mu + \sigma) = (5.3 - 8.372, 5.3 + 8.372) = (-3.072, 13.672)\).- Two standard deviations: \((\mu - 2\sigma, \mu + 2\sigma) = (5.3 - 2 \times 8.372, 5.3 + 2 \times 8.372) = (-11.444, 21.844)\).- Three standard deviations: \((\mu - 3\sigma, \mu + 3\sigma) = (5.3 - 3 \times 8.372, 5.3 + 3 \times 8.372) = (-19.816, 30.744)\)."

Step 3: Estimate Probability Using Empirical Rule

For part b, \((-11.444, 13.672)\) is from \(-1.5\sigma\) to \(+1\sigma\), roughly covering more than 68% but less than 95% of the data. A good approximation is 81.5%.

Step 4: Calculate Probability with Standard Normal Distribution

To find the exact probability in part c, convert \(x\) to \(z\)-scores:- For \(x = -11.444\), \(z = \frac{-11.444 - 5.3}{8.372} = -2\).- For \(x = 13.672\), \(z = \frac{13.672 - 5.3}{8.372} = 1\).Use the standard normal table or a calculator:- \(P(Z < 1) = 0.8413\) and \(P(Z < -2) = 0.0228\).- Hence, \(P(-2 < Z \leq 1) = 0.8413 - 0.0228 = 0.8185\) or 81.85%.

Key Concepts

Normal Distribution

The normal distribution is a fundamental concept in statistics, often referred to as a "bell curve" because of its distinctive shape. This distribution is symmetric around its mean, and most of the data points fall close to the mean.
Some key characteristics of a normal distribution include:

• Symmetry: The left and right sides of the curve are mirror images.
• The mean, median, and mode of the distribution are all equal.
• It is defined by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).

The normal distribution is crucial because many natural phenomena and measurement errors tend to approximate this shape. Understanding the properties of a normal distribution allows statisticians to make predictions about data and apply statistical tests.

Probability Density Function

A Probability Density Function (PDF) describes the likelihood of a continuous random variable taking on a particular value. For normal distributions, the PDF is shaped like a bell curve, centered around the mean (\(\mu\)) and determined by the standard deviation (\(\sigma\)).
The formula for the PDF of a normal distribution is:\[ f(x | \mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\]This formula allows us to calculate the probability of a random variable falling within a specified range, an essential aspect of probability theory.
The Empirical Rule, which states how data is distributed in a normal distribution, relies heavily on understanding the probability density function.

Standard Deviation

Standard deviation is a statistic that measures the dispersion of a dataset relative to its mean. In the context of the normal distribution, the standard deviation (\(\sigma\)) is key to understanding how spread out data are around the mean (\(\mu\)).
When data are close to the mean, we have a smaller standard deviation, indicating less spread. If data are spread out, the standard deviation will be larger.
Standard deviation is used in the Empirical Rule, which states:

• About 68% of data fall within one standard deviation (i.e., \(\mu \pm \sigma\)).
• Approximately 95% fall within two standard deviations (i.e., \(\mu \pm 2\sigma\)).
• Nearly 99.7% is within three standard deviations (i.e., \(\mu \pm 3\sigma\)).

Understanding standard deviation helps describe how typical or atypical a data point is within the distribution.

Z-Scores

Z-scores are a way of standardizing scores on a normal distribution. They measure how many standard deviations an element is from the mean.
The formula for calculating a z-score is:\[ z = \frac{x - \mu}{\sigma}\]This allows for comparison across different datasets by converting values to a common scale.
Using z-scores helps in analyzing data under a standard normal distribution (mean = 0 and standard deviation = 1), simplifying calculations like finding probabilities.
For example, when determining probabilities, z-scores can be used to find areas under the normal curve. Each z-score correlates to a percentile, which indicates the probability of a score falling below that z. Understanding z-scores can thus help in estimating the likelihood of various outcomes.