Problem 32 - An Introduction to Mathematical Statistics and Its Applications

Question

(a) Let \(00\). Which number is larger? \(\int_{a}^{a+1} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z\) or \(\int_{a-1 / 2}^{a+1 / 2} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z\)

Step-by-Step Solution

Answer

(a) The two numbers are equal. (b) \(\int_{a-1/2}^{a+1 / 2} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z > \int_{a}^{a+1} \frac{1}{\sqrt{2 \pi}} e^{-z^{2}/2} d z\)

Step 1: Compare the two integrals in (a)

The standard normal distribution is symmetric about \(z = 0\). Consequently, the two integrals \(\int_{a}^{b} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z \quad\) and \(\quad \int_{-b}^{-a} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z\) will yield identical results. The standard normal distribution's symmetry implies that the left hand side equals the right hand side..

Step 2: Compare the two integrals in (b)

Part (b) involves comparing \(\int_{a}^{a+1} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z\) and \(\int_{a-1/2}^{a+1 /2} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z\). The first integral is over the interval \(a\) to \(a+1\), meaning that the values of \(z\) range from \(a\) to \(a+1\). For the second integral, \(z\) ranges from \(a-1/2\) to \(a+1/2\). Because the standard normal distribution's graph is higher for values closer to \(z = 0\), and lower for values further away, the larger value will be the one where the integral is centered around 0. Therefore, \(\int_{a-1/2}^{a+1 / 2} \frac{1}{\sqrt{2 \pi}} e^{-z^{2}/2} d z > \int_{a}^{a+1} \frac{1}{\sqrt{2 \pi}} e^{-z^{2}/2} d z\), provided \(a>0\).

Key Concepts

Symmetric Distribution

In mathematics, a symmetric distribution is one where the left and right sides of the distribution are mirror images of each other. This concept plays a critical role in understanding the standard normal distribution, which is a common model for statistical analysis.

The standard normal distribution is an example of a symmetric distribution. It is symmetric about the vertical axis at zero (the mean). The bell-shaped curve rises to a peak at the center and falls away symmetrically on both sides. This symmetry is crucial because it allows us to predict probabilities and compare integrals easily.

For instance, when comparing two integrals across a symmetric distribution, such as \( \int_{a}^{b} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z \) and \( \int_{-b}^{-a} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z \), the symmetry ensures that both integrals represent the same area under the curve. Because these bounds are equidistant from the mean (zero), the areas are equal, demonstrating the powerful implication of symmetry in probability calculation.

Integral Comparison

Integral comparison is another important concept linked with the standard normal distribution and helps in comparing the probabilities over different intervals.

In the standard normal distribution, different intervals can be compared to analyze the likeliness of events. Specifically, if you need to compare two integrals like \( \int_{a}^{a+1} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z \) and \( \int_{a-1/2}^{a+1 / 2} \frac{1}{\sqrt{2 \pi}} e^{-z^{2}/2} d z \), the one with the greater value indicates a higher concentration of probability.

The reasoning here is based on the shape of the standard normal distribution. The peak of this distribution is at zero, meaning it has higher values closer to zero. Therefore, when one of the intervals is more centered around zero, its integral will sum to a larger value than an interval of the same width but farther from zero. This comparison helps in understanding which range has a higher probability under the curve.

Probability Intervals

Probability intervals are portions of a probability distribution that are used to estimate the probability of a random variable falling within a particular range.

Consider a standard normal distribution: the key to probability intervals lies in understanding which portion or interval of the distribution curve you are examining. Each interval contains a segment of the total probability, which adds up to one for a full distribution.

To determine probabilities between different intervals, compare their respective integrals, as we have seen. For a clear example, in cases like \( \int_{a}^{a+1} \frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2} d z \), the range represents how likely a value between these points is to occur. By integrating over this range, you effectively calculate the probability of the variable landing within this interval. Recognizing the nuances of these intervals allows us to interpret how the variable behaves across the distribution, and where most values cluster in relation to the mean—providing insights into the variable's variability.