Question

The heights of U.S. men are normally distributed with a mean of 69 inches and a standard deviation of 3 inches. This means that the fraction of men with a height between \(a\) and \(b\) (with \(a

Short answer

Answer: Approximately 68.25% of U.S. men have a height between 65 and 73 inches.

Step-by-step solution

Understand the normal distribution formula

The normal distribution function is given by the integral $$\frac{1}{\sigma \sqrt{2 \pi}} \int_{a}^{b} e^{-(x-\mu)^{2} / (2\sigma^2)} d x$$ where \(\mu\) is the mean, \(\sigma\) is the standard deviation, \(a\) and \(b\) are the limits of height we want to find the fraction between, and \(e\) is the Euler's number. In our case, the mean (\(\mu\)) is 69 inches and the standard deviation (\(\sigma\)) is 3 inches.

Plug in the values

Plug in the values of \(\mu\) and \(\sigma\) into the normal distribution function: $$\frac{1}{3 \sqrt{2 \pi}} \int_{a}^{b} e^{-[(x-69) / 3]^{2} / 2} d x$$

Set the limits a and b

Let's assume we want to find the fraction of men with a height between \(a = 65\) inches and \(b = 73\) inches. Plug in the values into the integral: $$\frac{1}{3 \sqrt{2 \pi}} \int_{65}^{73} e^{-[(x-69) / 3]^{2} / 2} d x$$

Evaluate the integral

To evaluate this integral, we will need to use a standard normal table or a calculator with built-in integration functions. Let's assume we will use a calculator with built-in integration functions. The result of the integral will be approximately: $$\frac{1}{3 \sqrt{2 \pi}} \times 1.8278$$

Calculate the fraction

Now, multiply the result of the integral by the constant in front to obtain the fraction: $$ \frac{1.8278}{3 \sqrt{2 \pi}} \approx 0.6825$$ So, approximately 68.25% of U.S. men have a height between 65 and 73 inches.

Key concepts

Mean and Standard Deviation

The mean and standard deviation are key components in statistics, especially when dealing with normal distributions. The mean, often symbolized as \( \mu \), is the average value you would expect from a set of data. In the context of our exercise, the mean is 69 inches, representing the average height of U.S. men.

The standard deviation, denoted as \( \sigma \), measures the amount of variation or dispersion in a set of values. A smaller standard deviation indicates that the values are close to the mean, while a larger standard deviation suggests more spread out data. Here, the standard deviation is 3 inches, implying that the majority of men's heights cluster around 69 inches with typical variances being around 3 inches from the mean.

Understanding these two concepts helps in interpreting the normal distribution and offers insights into how much individual values differ from the average.

Integral Calculus

Integral calculus is a branch of mathematics that is often used to calculate areas under curves. It's a fundamental tool in the evaluation of normal distributions. When discussing a probability distribution, integral calculus allows us to determine the probability of a random variable falling within a certain range.

In our specific exercise, we use the integral:

• \( \int_{a}^{b} e^{-[(x-69)/3]^2/2} dx \)

This formula helps us calculate the fraction or percentage of values that lie within the specified limits (between \( a \) and \( b \)) in the context of height distribution. By integrating this function between two specified points, we get the probability that a value will fall between those two points. Remember that using integral calculus in this way transforms an abstract notion of distribution into a quantifiable probability measure.

Therefore, when you perform such calculations, you're essentially calculating the area under a segment of the normal curve between the specified range, yielding valuable insights about the statistical properties of the distribution in question.

Probability Density Function

The probability density function (PDF) is crucial in statistics for describing the likelihood of different outcomes in a continuous random variable. For a normal distribution, the PDF is characterized by its bell-shaped curve, known as the Gaussian function.

The formula for a normal distribution's PDF is:

• \( \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \)

Where:

• \( x \) is the variable
• \( \mu \) is the mean
• \( \sigma \) is the standard deviation

The PDF doesn't provide probabilities directly, because they don't sum to 1 as in discrete distributions. Instead, the area under the curve for a range gives the probability of the variable falling within that range.

This characteristic of PDFs is an essential concept when determining the likelihood of specific outcomes in a continuous data set and ties directly into how integrals calculate probabilities for intervals in continuous distributions.