Z-Scores and Normal Distribution Detailed Notes

Warm-Up Problems

Function Composition and Graphing:
- Given f(x) = x - 1 and g(x) = -2x + 4, find the graph representing f(g(x)).
- This involves substituting g(x) into f(x).
Function Composition:
- Given f(x) = x - 1 and g(x) = -2x^2 + 4, find g(f(x)).
- This involves substituting f(x) into g(x).
- g(f(x)) = -2(x-1)^2 + 4.
Comparing Unlikely Events Using Z-Scores:
- Compare the likelihood of two events using z-scores.
- a) Score of 84, class average of 75, standard deviation of 3.
- b) Score of 84, class average of 71, standard deviation of 5.
- Calculate z-scores for both scenarios to determine which is more unlikely.

Z-score Calculation

The Z-score represents the number of standard deviations 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
- For scenario a): z = \frac{84 - 75}{3} = \frac{9}{3} = 3
- For scenario b): z = \frac{84 - 71}{5} = \frac{13}{5} = 2.6
  Therefore, a test score of 84 in scenario a) is more unlikely since it has a higher z-score.

Normal Distribution and Percentage of Scores:
- Test scores are normally distributed with a mean of 550 and a standard deviation of 82.
- What percentage of test scores are between 632 and 714?
- First calculate the z-scores for the given boundaries and use the z-table to determine the percentages.

Z-Scores Explained

Z-scores indicate the number of standard deviations from the mean.
A z-score of 0 means the score is the same as the mean.
Positive z-scores indicate values above the mean, while negative z-scores indicate values below the mean.

Examples

Calculating Z-score:
- Temperature of 72 degrees, mean of 60, standard deviation of 7.
- z = \frac{72 - 60}{7} = \frac{12}{7} \approx 1.71
Calculating Temperature from Z-score:
- Z-score of -2.4, mean of 60 degrees, standard deviation of 7.
- x = z \cdot \sigma + \mu
- x = -2.4 \cdot 7 + 60 = -16.8 + 60 = 43.2

You Try

Calculating Mean:
- Temperature of 80, z-score of 2, standard deviation of 5.
- 80 = 2 \cdot 5 + \mu
- 80 = 10 + \mu
- \mu = 70
Calculating Standard Deviation:
- Temperature of 63, z-score of -2.1, average of 76.
- 63 = -2.1 \cdot \sigma + 76
- -13 = -2.1 \cdot \sigma
- \sigma = \frac{-13}{-2.1} = 6.19

Revisiting a Problem

History test: average of 75, standard deviation of 7, Garrett scored 87.
Science test: average of 84, standard deviation of 3, Garrett scored 88.
On which test did Garrett do better compared to his peers?

Calculating Garrett's Z-scores

History: z = \frac{87 - 75}{7} = \frac{12}{7} \approx 1.71
Science: z = \frac{88 - 84}{3} = \frac{4}{3} \approx 1.33

Garrett performed better on History test with respect to his peers.

SOL-Like Z-Score Questions

Bench Press Data:
- The table shows the amount students can bench press for four different classes.
- The data is normally distributed.
- The data present includes: Name, Mean, Standard Deviation and Student's z-score

Data:

Name	Mean for Class	Standard Deviation for Class	Student’s z- score
Jacob	120	5.2	3.1
Tim	135	3.1	1.9
Layla	118	6.3	2.1
Aram	127	2.9	2.5

Which of the four students can bench press the most?

To find the answer, we can use the z-score formula:
x = z \cdot \sigma + \mu
Where:

x is the bench press amount
z is the z-score
σ is the standard deviation
μ is the mean

Using this formula:

Jacob’s bench press: x = 3.1 \cdot 5.2 + 120 = 16.12 + 120 = 136.12
Tim’s bench press: x = 1.9 \cdot 3.1 + 135 = 5.89 + 135 = 140.89
Layla’s bench press: x = 2.1 \cdot 6.3 + 118 = 13.23 + 118 = 131.23
Aram’s bench press: x = 2.5 \cdot 2.9 + 127 = 7.25 + 127 = 134.25

Tim can bench press the most.

Z-Score Calculation:
- A dataset is normally distributed with a mean of 75.54 and a standard deviation of 6.52.
- An element in this set is 63.47. What is the z-score for 63.47? Round your answer to the nearest hundredth.
- z = \frac{63.47 - 75.54}{6.52} = \frac{-12.07}{6.52} \approx -1.85
Histogram and Standard Deviations:
- The normally distributed data on the pollution levels for 34 cities are summarized in a histogram.
- The mean amount of pollution for these cities is 32.5. The standard deviation of the data is 3.5.
- Identify each interval that may have data points within 1.8 standard deviations of the mean.
- Calculate the range around the mean:
- Lower bound: 32.5 - (1.8 \cdot 3.5) = 32.5 - 6.3 = 26.2
- Upper bound: 32.5 + (1.8 \cdot 3.5) = 32.5 + 6.3 = 38.8

Additional Z-Score Problems

LDL Cholesterol Levels:
- The graph provides the LDL cholesterol levels for 250,000 males in the US. The mean is 111.7 mg/dL, and the standard deviation is 21.3 mg/dL. Each region is one standard deviation wide.
- Which region(s) correspond to the data for 34,000 US males?
- Calculate the percentage of the population each region represents.
- Region-1 is 1 standard deviation below the mean, so 34% of total values fall into this bucket.
- 0.34 \cdot 250000 = 85000
- So Region-1 consists of 85000 males, then it can't correspond to 34000 males.
*Looking at the bar on the graph each region represents about 17% of the total population.
- 0.17 \cdot 250000 = 42500
- So these regions can't correspond to 34000 males.
Test Scores and Z-Scores:
- Test scores are normally distributed with a mean of 78 and a standard deviation of 6.2. Uday’s z-score is -1.2, and Zima’s z-score is 0.9. How many points higher did Zima score on the test over Uday? Round to the nearest hundredth.
  - x = z \cdot \sigma + \mu
- Uday’s score: x = -1.2 \cdot 6.2 + 78 = -7.44 + 78 = 70.56
- Zima’s score: x = 0.9 \cdot 6.2 + 78 = 5.58 + 78 = 83.58
- Difference: 83.58 - 70.56 = 13.02
Systolic Blood Pressure:
- Data collected from 323,800 adults shows the average (mean) systolic blood pressure is 125 mmHg. The standard deviation for systolic blood pressure is 15 mmHg. Which region(s) correspond to the data for 6800 US males?
- Similar to problem 5, analyze the distribution and sizes of the regions relative to the given population to determine the corresponding regions.