Math 1153 - 8 October 2026 - Section 5.4, Chapter 6

Overview

Mike Jacobsen completes Section 5.4 by modeling Rochester, New York, December snowfall with a normal distribution (mean 21.9 inches, standard deviation 6.5) and practicing TI-84 normal CDF and inverse norm calculations. A cork-manufacturing example reinforces interval probabilities, the complement rule, and right-tail percentiles; the class closes by previewing Chapter 6’s analysis of two quantitative variables through scatter plots and correlation.

Key takeaways

For normal CDF on a TI-84, enter lower bound, upper bound, mean, and standard deviation in that order; multiply the returned probability by 100 when the question asks for a percentage.
On an older TI-84 inverse norm screen that supports only left-tail area, convert a 1% right-tail target to 0.99; on a newer model, select right tail and enter 0.01.
A graph is a practical error check: the 1% highest-snowfall cutoff must exceed Rochester’s 21.9-inch mean, so a result near 6.8 inches signals the wrong tail.
Under the Rochester model (μ = 21.9 inches, σ = 6.5 inches), the chance of 10 inches or less is 3.36%, the chance of 10–20 inches is 35.15%, and the 1% upper-tail cutoff is about 37.0 inches.
For corks with μ = 3 cm and σ = 0.1 cm, only 31.08% fall within the 2.96–3.04 cm specification; the complement rule gives 68.92% defective by size.
The 1% largest cork diameters begin at approximately 3.23 cm, found with inverse norm using a 0.01 right-tail area or equivalent 0.99 left-tail area.

Chapters

0:00 Quiz 6, Exam 2, and the Section 5.4 Plan

Quiz 5 is due, and Quiz 6 covers Section 5.4 normal-distribution problems using calculator percentages and inverse norm.
Exam 2 will cover material through Section 5.4; the TI-84 is vital for the calculator-based methods.
Mike Jacobsen plans to post sample exam notes and scheduling information, following the process used for Exam 1.

1:30 Exam 2 Scope and Midterm-Week Logistics

The corrected Exam 2 scope excludes Chapter 1 and reaches through Section 5.4.
There will be no separate midterm exam; grades will be reported at the end of midterm week.
Exam scheduling information is expected at the start of the following week.

3:30 Rochester Snowfall Model: Mean 21.9 Inches, SD 6.5

December snowfall in Rochester, New York, is modeled as normally distributed with mean μ = 21.9 inches and standard deviation σ = 6.5 inches.
The 6.5-inch standard deviation represents substantial variability; two standard deviations span about 13 inches from the mean.
These population parameters are recorded first to support the later calculator work and partial credit.

5:00 TI-84 Normal CDF for 10 Inches of Snow or Less

The question asks for a percentage and specifies 10 inches or less, indicating normal CDF and left-tail shading.
On the TI-84, open the distribution menu with 2nd and VARS, then select normal CDF.
Enter lower bound −10^99, upper bound 10, mean 21.9, and standard deviation 6.5, in that order.
The calculator returns about 0.033567; multiply by 100 to report a 3.36% chance.

12:00 Recovering the TI-84 Screen and Checking the CDF Result

After a student cannot see the calculator display, Mike Jacobsen restarts screen sharing and repeats the TI-84 input sequence.
On the result screen, repeated Enter presses display the probability; multiplying the result by 100 converts it to a percentage.
The 3.36% result means snowfall of 10 inches or less is rare under the Rochester model.

15:42 Normal CDF for Rochester Snowfall Between 10 and 20 Inches

“Between 10 and 20 inches” calls for shading only the interval between those bounds.
Use normal CDF with lower bound 10, upper bound 20, μ = 21.9, and σ = 6.5.
The calculator gives approximately 0.3515, or a 35.15% chance of snowfall in that range.
A rough graph and a probability sanity check help identify results that are implausibly high or low.

21:00 Interpreting the 1% of Worst Rochester Snowfall Cases

The question supplies a percentage and asks for a snowfall amount, so the required operation is inverse norm.
“Worst cases” is interpreted as unusually high snowfall, meaning the desired percentile lies in the right tail.
The unknown amount is labeled K and placed to the right of the mean 21.9 inches, with 1% of the area beyond it.

24:00 Inverse Norm Tail Settings and the 37-Inch Threshold

Older TI-84 models accept only left-tail area, so a 1% right tail is entered as left-tail area 0.99.
Newer TI-84 models with tail selectors should use right tail and area 0.01; either setup targets the same cutoff.
Enter μ = 21.9 and σ = 6.5 after the area; inverse norm gives about 37.0 inches for the 1% highest-snowfall threshold.
A mistaken left-tail area of 0.01 would produce a low value near 6.8 inches, contradicting the graph’s right-tail location.

36:50 Calculator Troubleshooting and Choosing More Section 5.4 Practice

For a TI-84 Plus CE displaying “MathPrint Classic,” press 2nd and MODE to leave the screen.
The class chooses another Section 5.4 example rather than beginning Chapter 6 with only a few minutes remaining.
Mike Jacobsen posts an extra example to Canvas as a Word document, noting that its question format resembles quiz and exam problems.

40:10 Cork-Machine Parameters: Mean 3 cm, SD 0.1 cm

A machine produces wine corks whose diameters are modeled by a normal distribution with mean 3 cm and standard deviation 0.1 cm.
The correct population-parameter symbols are μ = 3 and σ = 0.1; Y-bar and s would describe sample statistics instead.
The cork diameter must match a bottle specification, making variability important even when the machine’s average is correct.

44:30 Cork Specification Probability Using Normal CDF

The bottle requires diameters between 2.96 cm and 3.04 cm, so the target is the area between those bounds.
Use TI-84 normal CDF with lower bound 2.96, upper bound 3.04, mean 3, and standard deviation 0.1.
The resulting probability is about 0.3108, or 31.08% of corks within specification.
The interval is narrower than one standard deviation on either side of the mean, so a result well below 68% is reasonable.

50:50 The Complement Rule Finds the Defective Cork Percentage

Corks outside 2.96–3.04 cm are treated as defective due to incorrect size.
Rather than run another normal CDF calculation, subtract the in-specification rate from the whole: 100% − 31.08% = 68.92%.
This complement-rule result shows that the modeled machine produces a large share of unusable corks despite its 3 cm average.

55:10 Right-Tail Inverse Norm for Corks and Preview of Chapter 6

The 1% of largest cork diameters requires inverse norm and right-tail shading, with the target cutoff K above the 3 cm mean.
Enter area 0.01 with a right-tail selector on newer TI-84 models; older models use left-tail area 0.99, with μ = 3 and σ = 0.1.
Inverse norm gives approximately 3.23 cm, so corks at or above that diameter comprise the largest 1%.
Chapter 6 will study two quantitative variables using scatter plots, association, and correlation; class ends before starting the new material.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Mike Jacobsen.