Problem 11 - Calculus for Biology and Medicine

Question

Denote by the density of a normal distribution with mean \(\mu\) and standard deviation \(\sigma\) $$ f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2 \sigma^{2}} $$ for \(-\infty

Step-by-Step Solution

Answer

(a) \(f(x)\) is symmetric about \(x=\mu\). (b) Maximum at \(x=\mu\). (c) Inflection points at \(x=\mu\pm\sigma\). (d) Graph is a bell curve with peak at \(x=2\).

Step 1: Understand the Function

The density function of a normal distribution is given by \[ f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}}. \]This function is defined for all real values of \(x\).

Step 2: Symmetry about x=μ

Substituting \(x = a\) we have,\[ f(a) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(a-\mu)^2}{2 \sigma^2}}. \]Then substitute \(x = 2\mu - a\), \[ f(2\mu - a) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{((2\mu-a)-\mu)^2}{2 \sigma^2}} = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(a-\mu)^2}{2 \sigma^2}} = f(a). \]The function is symmetric about \(x = \mu\).

Step 3: Find the Maximum of f(x) at x=μ

The maximum of a symmetric function occurs at the axis of symmetry. To confirm:Derive the function with respect to \(x\):\[ f'(x) = f(x) \cdot \left(-\frac{(x-\mu)}{\sigma^2}\right) = 0 \]Solving, we find \(x = \mu\). Thus, the maximum of \(f(x)\) is at \(x = \mu\).

Step 4: Inflection Points

An inflection point occurs where the second derivative is zero and changes sign. Find the second derivative, \[ f''(x) = \frac{f'(x) \cdot (x-\mu) - f(x)}{\sigma^2} = 0. \]Solve \[ x - \mu = \pm \sigma. \]So, the inflection points are at \(x = \mu - \sigma\) and \(x = \mu + \sigma\).

Step 5: Graph for μ=2, σ=1

For \(\mu = 2\) and \(\sigma = 1\), the function becomes \[ f(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{(x-2)^2}{2}}. \]Graph this function in the real number plane for \(x\). It should show a bell curve with its peak at \(x = 2\) and inflections at \(x = 1\) and \(x = 3\).

Key Concepts

Mean and Standard Deviation

The mean \(\mu\) and standard deviation \(\sigma\) are crucial in understanding a normal distribution. The mean represents the average or central value of the distribution. It’s where the peak of the bell curve is centered. Think of \(\mu\) as the point at which the distribution is balanced.
On the other hand, the standard deviation is a measure of the spread or width of the distribution. A smaller \(\sigma\) results in a steeper curve, while a larger \(\sigma\) stretches the curve out, making it flatter.
The values around the mean within one standard deviation capture about 68% of the distribution’s probability, illustrating how tightly or loosely data is spread around the mean.

Density Function

The density function of a normal distribution is represented mathematically as \( f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x-\mu)^2}{2 \sigma^2}} \). This function tells us how likely different values of \(x\) are, based on the mean and standard deviation.
The exponential term \(e^{-\frac{(x-\mu)^2}{2 \sigma^2}}\) ensures that values closer to the mean \(\mu\) are more likely, reflected by the higher peak on the density curve. The factor \(\frac{1}{\sigma \sqrt{2 \pi}}\) normalizes the area under the curve to 1, representing a complete probability distribution.

Inflection Points

Inflection points in the context of the normal distribution are the points where the concavity of the curve changes. They are located at \( x = \mu - \sigma \) and \( x = \mu + \sigma \).
An inflection point is where the curve changes from being concave (curving upwards) to convex (curving downwards) or vice versa. This shift indicates the bounds within which the data is spread more densely or sparsely.
In simpler terms, the inflection points signify where the slope of the curve starts decreasing less rapidly and mark the transition between the steepest parts of the curve and the flatter tails.

Maximum Value

The maximum value of the normal distribution density function is located at \( x = \mu \), which is the mean of the distribution. At this point, the curve reaches its peak, demonstrating the highest probability density.
This maximum occurs because the symmetry of the function ensures that all values equidistant from the mean have the same probability density, leading to \( x = \mu \) being the most probable or typical value in the distribution.
The height of the peak itself depends on the standard deviation \(\sigma\); a smaller \(\sigma\) leads to a higher peak, showcasing the key influence of the standard deviation on the distribution's shape.

Symmetry About the Mean

A normal distribution is characterized by its perfect symmetry around the mean \(\mu\). This symmetry implies that the left side of the distribution is a mirror image of the right side.
Mathematically, this is proven by substituting symmetric points \(x = a\) and \(x = 2\mu - a\), which yield the same density value \( f(a) = f(2\mu - a) \).
This feature of symmetry about the mean \(\mu\) is crucial in simplifying the analysis and interpretation of normal data, as it allows for straightforward predictions about the probability of data lying within certain intervals around the mean.