Using the normal approximation to the binomial distribution, and tables [or calculator for \(\phi(t)\) ], find the approximate probability of each of the following: Find the probabilities for a normally distributed random variable to differ by more than \(\sigma\), \(2 \sigma, 3 \sigma, 4 \sigma\), from its mean value. Your answers should satisfy Chebyshev's inequality (for example, the probability of a deviation of more than \(2 \sigma\) is less than \(\frac{1}{4}\) ). You will find, however, that for the normal distribution, the probabilities are actually much smaller than Chebyshev's inequality requires.

Chebyshev's inequality is a fundamental concept in probability theory. It provides an upper bound for the probability that the value of a random variable with a given mean and variance lies within a certain number of standard deviations from the mean. This theorem is particularly useful because it applies to all distributions, regardless of their shape.

Mathematically, Chebyshev's inequality states that for any random variable X with mean \(\backslashmu\) and standard deviation \(\backslashsigma\), the probability that X is more than k standard deviations away from the mean is at most \(\frac{1}{k^2}\).

This can be expressed as:

\[ P(\mid X - \mu\mid \geq k \sigma) \leq \frac{1}{k^2} \]

For example, the probability that X deviates by more than 2 standard deviations from the mean is at most \(\frac{1}{4}\) or 0.25. This provides a conservative estimate since the actual probabilities for specific distributions, such as the normal distribution, are usually much smaller.

The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. Using this standardized form simplifies the calculation of probabilities and helps in comparing different normal distributions.

The probability density function (PDF) of the standard normal distribution is given by:

\[ f(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \]

When working with normally distributed data, converting it to the standard normal form is often helpful. This is done using z-scores.

A z-score measures how many standard deviations a data point is from the mean. If you have a normally distributed variable X with mean \(\mu\) and standard deviation \(\sigma\), the z-score for a value x is calculated as:

\( Z = \frac{x - \mu}{\sigma} \)

Z-scores are useful because they convert data into a standard form, allowing the use of standard normal distribution tables to find probabilities.

In the context of the example, the differences of \(\sigma\), \(2\sigma\), \(3\sigma\), and \(4\sigma\) were converted into z-scores of 1, 2, 3, and 4 respectively. These z-scores help in determining the probabilities using the standard normal distribution.

Probability calculations using the standard normal distribution involve finding areas under the curve. This can be done using z-tables or statistical software.

For each z-score calculated from the exercise, we look up the area under the curve to the left of that z-score in a standard normal table. For example, for a z-score of 1:

\( P(Z < 1) = 0.8413 \)

To find the probability that the variable differs by more than the z-score value, we calculate:

\( P(Z > 1) = 1 - P(Z < 1) = 1 - 0.8413 = 0.1587 \)

Since the question asks for the probability of differing by more or less than the z-score, we multiply this value by 2:

\( P(\mid Z \mid > 1) = 2 \times 0.1587 = 0.3174 \)

This method is applied to the other z-scores (2, 3, 4) to find their probabilities, demonstrating that the actual probabilities are much smaller than those given by Chebyshev's inequality.