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Z-score conversion is a process used to transform an individual score from a normal distribution into a standardized form. This is especially helpful when comparing scores from different normal distributions or calculating probabilities in a standard normal distribution, which has a mean of 0 and a standard deviation of 1. To find the z-score, use the formula:\[ z = \frac{x - \mu}{\sigma}\]Where:

• The score (\(x\)) is the value you want to convert.
• The mean (\(\mu\)) is the average of all scores in the distribution.
• The standard deviation (\(\sigma\)) measures how spread out the scores are.

In our example, the test scores are normally distributed with \(\mu = 100\) and \(\sigma = 10\). The score of 90 becomes \(z = \frac{90 - 100}{10} = -1\), and the score of 110 becomes \(z = \frac{110 - 100}{10} = 1\). By converting these scores, we can now find probabilities using the standard normal distribution.

The Maclaurin series is a method of approximating functions using a polynomial expansion. Especially useful when evaluating functions that are difficult to integrate directly, it consists of infinite terms derived from the function's derivatives at the point where the expansion is centered. For example, the Maclaurin series for \( e^{-x^2/2} \), expanded to degree 10, is:\[ 1 - \frac{x^2}{2} + \frac{x^4}{8} - \frac{x^6}{48} + \frac{x^8}{384} - \frac{x^{10}}{3840} \] Here’s why this series is useful:

• Provides an easier way to approximate the function in a given range.
• Helps estimate the probability when combined with integration.
• The degree 10 polynomial gives a detailed enough approximation for many practical purposes.

Multiply this series by \( \frac{1}{\sqrt{2\pi}} \) to align it with the standard normal density curve. This conversion simplifies the integration process used to find probabilities.

Estimating probabilities under a normal distribution involves finding the area under the curve between two values. By converting test scores into z-scores, we can use the standard normal distribution to find these probabilities. The probability that a score lies between two points is represented by the integral of the normal density function over that interval. For values from 90 to 110 in the normal distribution with mean 100 and standard deviation 10, the corresponding z-scores are -1 and 1. Therefore, we estimate the probability by calculating:\[ P(90 \leq X \leq 110) = \int_{-1}^{1} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx \]This integral measures the area under the curve from \(x = -1\) to \(x = 1\). Use the Maclaurin series to approximate the value of this integral, providing an estimate of the probability, which ends up to be approximately 0.6827. This means about 68.27% of scores fall between 90 and 110.

Definite integration is a powerful mathematical tool for finding areas under curves, particularly useful in the context of probability. It allows calculation of the exact numerical probability of an event occurring within a specified range. In the context of normal distribution, when we calculate: \[ \int_{a}^{b} f(x) \, dx \]Here, \(f(x)\) represents the probability density function of the normal distribution, and the limits \(a\) and \(b\) correspond to desired score boundaries. When approximating using the Maclaurin series, term-by-term integration over the interval [-1, 1] gives:

• \( \int_{-1}^{1} \frac{1}{\sqrt{2\pi}} \, dx = \frac{2}{\sqrt{2\pi}} \)
• \( \int_{-1}^{1} -\frac{1}{2\sqrt{2\pi}} x^2 \, dx = 0 \)
• \( \int_{-1}^{1} \frac{1}{8\sqrt{2\pi}} x^4 \, dx = \frac{2}{15\sqrt{2\pi}} \)
• \( \int_{-1}^{1} -\frac{1}{48\sqrt{2\pi}} x^6 \, dx = -\frac{2}{315\sqrt{2\pi}} \)
• \( \int_{-1}^{1} \frac{1}{384\sqrt{2\pi}} x^8 \, dx = \frac{2}{2835\sqrt{2\pi}} \)
• \( \int_{-1}^{1} -\frac{1}{3840\sqrt{2\pi}} x^{10} \, dx = -\frac{2}{31185\sqrt{2\pi}} \)

Summing these results, we estimate the probability that scores fall within our specified boundaries, which, as an approximation, shows up as a specific numerical value.