Fall 2024: Sections 8-4, 8-5, and 8-6

Welcome to the Fall 2024 semester! This post provides a review of the material covered in sections 8-4, 8-5, and 8-6 of our textbook. We'll explore applications of the normal distribution and how it can be used to approximate the binomial distribution. Let's dive in!

8.4: Applications of the Normal Distribution

The normal distribution is a cornerstone of statistics. Remember, a normal distribution is defined by its mean, $\mu$, and standard deviation, $\sigma$. A critical concept here is the Z-score, which tells us how many standard deviations away from the mean a particular value is. The formula for calculating the Z-score is:

$$Z = \frac{X - \mu}{\sigma}$$

Where:

  • $X$ is the value you are interested in.
  • $\mu$ is the mean of the distribution.
  • $\sigma$ is the standard deviation of the distribution.

Example: Suppose a national testing service gives a test where scores are normally distributed with a mean of 400 and a standard deviation of 100. If you score 644, what fraction of students exceeded your score? First, calculate the Z-score:

$$Z = \frac{644 - 400}{100} = 2.44$$

Looking up this Z-score in a standard normal table gives us a value of approximately 0.9927. This means that 99.27% of the scores are *below* 644. Therefore, only $1 - 0.9927 = 0.0073$, or 0.73%, of students exceeded your score. Great job!

Example: If the body temperatures of adults are normally distributed with a mean of 98.6°F and a standard deviation of 0.73°F, what temperature represents the 90th percentile?

We know that $Z=1.28$ is the z-score for the 90th percentile. Using the equation $X = Z(\sigma) + \mu$, we can plug in the values:

$$X = 1.28(0.73) + 98.6 = 99.53$$

Therefore, 99.53°F represents the 90th percentile.

8.6: Approximation to the Binomial Distribution

The binomial distribution deals with the probability of success or failure in a series of independent trials. However, when the number of trials, $n$, is large, calculating binomial probabilities can be cumbersome. In these cases, we can use the normal distribution to *approximate* the binomial distribution. We can approximate binomial distribution to normal distribution if $np \geq 5$ and $nq \geq 5$ where $q = 1-p$.

For a binomial distribution with $n$ trials and probability of success $p$, the mean and standard deviation are:

  • Mean: $\mu = np$
  • Standard Deviation: $\sigma = \sqrt{np(1-p)}$

Example: A small airline knows that 20% of people with reservations don't show up. If a commuter plane holds 15 people and they overbook to 18, what's the probability more than 15 people will show up? Here, $n = 18$ and $p = 0.80$ (probability of showing up). Then, $\mu = 18 * 0.80 = 14.4$ and $\sigma = \sqrt{18 * 0.80 * 0.20} \approx 1.70$. To find the probability, we need to calculate $P(X > 15)$, and apply continuity correction by calculating the Z-score with $X = 15.5$

$$Z = \frac{15.5 - 14.4}{1.70} \approx 0.65$$

Looking up $Z = 0.65$ will give approximately $0.7422$, so $P(X > 15) = 1 - 0.7422 \approx 0.2578$.

Keep practicing these concepts, and you'll be well-prepared! Good luck with your studies!