Class Notes: March 23, 2023 - Sections 6-3 and 6-5

Welcome back to Professor Baker's Math Class! Today, we covered key concepts from Sections 6-3 and 6-5, focusing on normal distributions and approximating binomial probabilities.

Understanding Normal Distributions

The normal distribution is a cornerstone of statistics. Let's review how to determine percentages and probabilities for normally distributed variables.

  1. Sketch the normal curve: Always start by visualizing the distribution. This helps in understanding the problem.
  2. Shade the region of interest: Identify and shade the area that represents the probability you're trying to find. Mark the delimiting x-values.
  3. Calculate the z-score(s): The z-score tells you how many standard deviations away from the mean a particular value is. The formula is: $$z = \frac{x - \mu}{\sigma}$$ where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.
  4. Use Table II (or a calculator) to find the area: Table II provides the area under the standard normal curve corresponding to the calculated z-score. This area represents the probability.

Example: Intelligence Quotients (IQs)

Let's consider an example: Intelligence quotients (IQs) are normally distributed with a mean ($\mu$) of 100 and a standard deviation ($\sigma$) of 16. What percentage of people have IQs between 115 and 140?

  1. First, calculate the z-scores for 115 and 140: $$z_{115} = \frac{115 - 100}{16} = 0.94$$ $$z_{140} = \frac{140 - 100}{16} = 2.5$$
  2. Next, use Table II to find the corresponding areas. For $z = 0.94$, the area is 0.8264. For $z = 2.5$, the area is 0.9938.
  3. Finally, subtract the smaller area from the larger area: $0.9938 - 0.8264 = 0.1674$. Therefore, approximately 16.74% of people have IQs between 115 and 140.

Approximating Binomial Probabilities with Normal Curves

When dealing with a large number of trials in a binomial distribution, approximating probabilities using the normal curve can simplify calculations.

  1. Verify Conditions: Ensure that $np$ and $n(1-p)$ are both greater than or equal to 5. Here, $n$ is the number of trials and $p$ is the probability of success.
  2. Calculate Mean and Standard Deviation: The mean ($\mu$) is given by $\mu = np$, and the standard deviation ($\sigma$) is given by $\sigma = \sqrt{np(1-p)}$.
  3. Continuity Correction: Since we are approximating a discrete distribution (binomial) with a continuous distribution (normal), we use a continuity correction. For example, to find the probability of *exactly* 390 successes, we consider the interval from 389.5 to 390.5.
  4. Calculate Z-scores and Probabilities: Calculate the z-scores for the corrected interval endpoints and find the corresponding areas under the normal curve.

Example: Mortality

Suppose the probability of a person of age 20 being alive at age 65 is 0.80. If we randomly select 500 people of age 20, what is the probability that exactly 390 of them will be alive at age 65?

  1. Here, $n = 500$ and $p = 0.80$. So, $\mu = 500 * 0.80 = 400$ and $\sigma = \sqrt{500 * 0.80 * 0.20} = 8.94$.
  2. Using the continuity correction, we want the probability between 389.5 and 390.5. The z-scores are: $$z_{389.5} = \frac{389.5 - 400}{8.94} = -1.17$$ $$z_{390.5} = \frac{390.5 - 400}{8.94} = -1.06$$
  3. Consulting the z-table, the areas corresponding to these z-scores are approximately 0.1210 and 0.1446. Subtracting, we get $0.1446 - 0.1210 = 0.0236$. Therefore, the probability is approximately 2.36%.

Keep practicing these concepts, and you'll master normal distributions and binomial approximations in no time! See you in the next class!