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Find the expected values and the standard deviations (by inspection) of the normal random variables with the density functions given. $$\frac{1}{5 \sqrt{2 \pi}} e^{-(1 / 2)((x-3) / 5)^{2}}$$

Short Answer

Expert verified
Expected value: 3, Standard deviation: 5

Step by step solution

01

Identify the Structure of the Normal Density Function

The given density function is ewline ewline \(\frac{1}{5 \sqrt{2 \pi}} e^{-\frac{1}{2} \left(\frac{x-3}{5}\right)^2}\).ewline ewline This can be compared to the general form of a normal density function: ewline ewline \(\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2} \(\frac{x - \mu}{\sigma}\)^2} \).ewline We'll extract the values of \( \mu \) and \( \sigma \) by comparing coefficients.
02

Determine the Mean (\( \mu \))

By comparing the given density function with the standard form, it is clear that \( \mu \) is the value that shifts the distribution along the x-axis. In the provided function, \( x - 3 \) shifts the distribution, so \( \mu = 3 \).
03

Determine the Standard Deviation (\( \sigma \))

The value of \( \sigma \) is the denominator of the fraction inside the exponential function that involves \( x \). By comparing, we see that the denominator here is 5; thus, \( \sigma = 5 \).
04

State the Expected Value and Standard Deviation

From the previous steps, the expected value (mean), \( \mu \), and the standard deviation, \( \sigma \), have been found. The expected value is 3 and the standard deviation is 5.

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Key Concepts

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

Expected Value
The expected value, also known as the mean (\mu), is a measure of the central tendency of a probability distribution. In the context of a normal distribution, it tells us where the 'center' or 'peak' of the distribution is located. For the given density function, \[\frac{1}{5 \sqrt{2 \pi}} e^{-\frac{1}{2} \(\frac{x-3}{5}\)^2}\], we identify the expected value by comparing the function to the standard form of a normal distribution's density function: \[\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2} \(\frac{x-\mu}{\sigma}\)^2}\].
The term \(x - 3\) within the exponent indicates that the entire distribution is shifted by 3 units to the right along the x-axis. Therefore, the expected value \mu is 3. This means that, on average, you would expect a randomly chosen value from this distribution to be around 3. This concept is crucial for understanding how data behaves around the mean in many practical scenarios, like quality control and risk assessment.
Standard Deviation
The standard deviation (\sigma) is a measure of the amount of variation or dispersion in a set of values. For a normal distribution, it indicates how much the values typically deviate from the mean. In the given function, the standard deviation is identified by the term in the denominator within the exponent: \[\frac{1}{5 \sqrt{2 \pi}} e^{-\frac{1}{2} \(\frac{x-3}{5}\)^2}\].
By comparing it to the standard normal density function \[\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2} \(\frac{x-\mu}{\sigma}\)^2}\], we see that \(\text{denominator} = 5\.\) Hence, the standard deviation \sigma is 5. In practical terms, a larger standard deviation indicates the data points are more spread out from the mean, whereas a smaller standard deviation indicates they are closer to the mean. This statistic is used in fields ranging from finance (to measure market volatility) to engineering (to assess product consistency).
Density Function
A density function describes the probability distribution of a continuous random variable. For a normal distribution, the density function is given by: \[\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2} \(\frac{x-\mu}{\sigma}\)^2}\].
The density function provides important information about how values are distributed. The above form shows how the values are most likely to occur around the mean (expected value) \mu, and the role of the standard deviation \sigma. The exponent part \(-\frac{1}{2} \(\frac{x-\mu}{\sigma}\)^2\) signifies that values close to \mu have higher densities, making them more probable. In more detailed terms, as you move away from \mu by more than one standard deviation, the probability of those values decreases exponentially. For practical applications, this function helps in understanding probabilities within different ranges. For example, in assessing probabilities for outcomes in different sectors like investing, manufacturing, and various fields requiring statistical analysis, the density function is vital.

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Most popular questions from this chapter

In a certain cell population, cells divide every 10 days, and the age of a cell selected at random is a random variable \(X\) with the density function \(f(x)=2 k e^{-k x}, 0 \leq x \leq 10, k=(\ln 2) / 10.\) Find the probability that a cell is at most 5 days old.

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