Chapter 6-8 Test Review

Hello Math Students! This review will help you prepare for the upcoming test covering Chapters 6-8. Let's dive into some key concepts and examples to solidify your understanding.

Normal Distributions

Remember that normal distributions are essential for understanding many real-world phenomena. A key concept is the Z-score, which tells you how many standard deviations a particular value is from the mean. The formula for calculating the Z-score is:

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

Where:

  • $x$ is the observed value
  • $\mu$ is the mean of the distribution
  • $\sigma$ is the standard deviation of the distribution

Example 1: The lengths of nails produced in a factory are normally distributed with a mean of 4.81 cm and a standard deviation of 0.05 cm. We want to find the lengths that separate the top and bottom 10%. Since this involves finding x-values corresponding to certain probabilities (or areas under the normal curve), we can use the Z-score formula, but solve for x. The z-score for the top 10% is approximately 1.28 and for the bottom 10% is -1.28. First, let's calculate the length for the bottom 10%:

$$ x = Z\sigma + \mu = -1.28 * 0.05 + 4.81 = 4.746$$

Rounding to the nearest hundredth, we get 4.75 cm.

Next, we find the length for the top 10%:

$$ x = Z\sigma + \mu = 1.28 * 0.05 + 4.81 = 4.874$$

Rounding to the nearest hundredth, we get 4.87 cm.

Example 2: Trucks in a delivery fleet travel a mean of 80 miles per day with a standard deviation of 30 miles per day. What's the probability a truck drives between 97 and 107 miles? Use the z-score formula to calculate the z-scores and then find the corresponding probabilities using a z-table:

For 97 miles: $Z = \frac{97 - 80}{30} = 0.57$

For 107 miles: $Z = \frac{107 - 80}{30} = 0.90$

Using a Z-table, P(Z < 0.57) ≈ 0.7157 and P(Z < 0.90) ≈ 0.8159. Thus, the probability the trucks drives between 97 and 107 miles is approximately 0.8159 - 0.7157 = 0.1002.

Probability and Combinations

Example 3: A newspaper company classifies its customers by gender and residence. Given the provided data, what is the probability that a customer is *not male* or does *not live in a dorm*? First, let's calculate the number of customers that are not male, the number that do not live in a dorm, and the number that satisfies either condition. From the provided tables, the number of female customers is 227+254+165+219+283 = 1148. The number of customers who do not live in a dorm is 1870 - 129 - 254 = 1487. The total customers who are not male, or do not live in a dorm is 1148+ 593 = 1741.

So, the probability is $\frac{1741}{1870} ≈ 0.9310$

Example 4: There are 10 people in an office and 3 different phone lines. If all lines start ringing at once, how many groups of 3 people can answer the lines?

Since the order doesn't matter (any group of 3 can answer the phone lines) this is a combination problem. We want to calculate "10 choose 3", which is ${10 \choose 3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10*9*8}{3*2*1} = 120$

Discrete Probability Distributions

Remember to understand how to calculate expected value and variance for discrete random variables.

Given a discrete random variable $X$ and its probability mass function $P(X=x)$, the expected value is:

$$E(X) = \sum [x \cdot P(x)]$$

The variance is:

$$V(X) = \sum [(x - \mu)^2 \cdot P(x)]$$

Where $\mu = E(X)$ is the expected value.

Good luck with your test! Remember to review your notes, practice problems, and come prepared to show what you've learned! You've got this!