07 Jul Probability Homework
11. Sally Monilla has started serving lunch at her new business called Salmonilla’s Diner. During lunch hours, they serve an average of 10 customers per hour. What is the probability of serving less than three customers in any 15 minute period during lunch hours?
- Johnny Ray has 1000 egg-laying chickens at Eggstravagant Poultry Company.
Historically, each chicken lays an average of 260 eggs per year or 5 per week, per chicken. That comes to an average of 5000 eggs per week or 260,000 total eggs laid per year, which can be considered a steady stream of egg production. Historically, 15% of the eggs are damaged before Johnny Ray can gather them.
a) If 20 eggs are sampled, how many would you expect to be damaged?
b) What is the probability that a sample of 20 eggs produced will have 3 or more damaged?
Use the data in the table below to answer problems 13 and 14
To receive full credit for the problems below, you must show:
*The equations you used to arrive at your solution.
*The numbers you substituted into those equations.
| Subgroup | Data 1 | Data 2 | Data 3 | Data 4 | Data 5 | Comments | Average | Range |
| 1 | 84.7 | 84.8 | 85.2 | 85.1 | 85.0 | 84.96 | 0.50 | |
| 2 | 85.4 | 85.0 | 84.6 | 84.9 | 85.2 | 85.02 | 0.80 | |
| 3 | 84.9 | 85.0 | 85.1 | 85.1 | 84.9 | 85.00 | 0.20 | |
| 4 | 84.1 | 84.2 | 84.0 | 84.3 | 84.0 | New Employee | 84.12 | 0.30 |
| 5 | 85.1 | 84.8 | 85.0 | 84.9 | 84.8 | 84.92 | 0.30 | |
| 6 | 85.2 | 84.8 | 84.9 | 85.1 | 85.1 | 85.02 | 0.40 | |
| 7 | 85.3 | 85.4 | 85.1 | 85.0 | 85.2 | Just Lubricated | 85.20 | 0.40 |
| 8 | 85.1 | 84.9 | 85.2 | 85.1 | 85.0 | 85.06 | 0.30 | |
| 9 | 84.5 | 84.8 | 84.6 | 84.7 | 84.5 | New lot of raw material | 84.62 | 0.30 |
| 10 | 84.8 | 85.1 | 85.3 | 85.5 | 85.2 | 85.18 | 0.70 | |
| 11 | 84.9 | 84.8 | 85.0 | 85.1 | 85.2 | 85.00 | 0.40 | |
| 12 | 85.3 | 85.0 | 85.2 | 84.7 | 84.9 | 85.02 | 0.60 | |
| 13 | 84.6 | 84.7 | 84.9 | 84.8 | 84.7 | 84.74 | 0.30 | |
| 14 | 85.0 | 85.1 | 85.0 | 85.0 | 85.1 | Different Materials | 85.04 | 0.10 |
| 15 | 85.2 | 84.9 | 85.0 | 85.1 | 84.9 | 85.02 | 0.30 | |
| 16 | 85.3 | 84.9 | 85.2 | 84.9 | 85.0 | 85.06 | 0.40 | |
| 17 | 84.9 | 85.7 | 85.6 | 84.2 | 84.3 | New Tool Installed | 84.94 | 1.50 |
| 18 | 85.3 | 85.1 | 84.7 | 84.9 | 85.0 | 85.00 | 0.60 | |
| 19 | 85.1 | 85.2 | 85.3 | 84.9 | 85.1 | 85.12 | 0.40 | |
| 20 | 84.8 | 84.9 | 85.2 | 85.0 | 85.1 | 85.00 | 0.40 | |
| Average | 84.95 | 0.46 |
13: Use the information in the above table to compute the Control Limits for an Average and a Range Control Chart.
14: Do NOT assume assignable cause variation. If there are any subgroups out-of-control, use the “Comments” column to determine if assignable cause variation is present. Discard any subgroups that are out of control and have assignable cause variation and calculate revised Control Limits for an Average and a Range Control Chart. Only look at whether they are outside the control limit- not all the zone rules. Be sure to list any subgroups that are removed.
Subgroups removed:_________________________
15. What is your assessment of the process after your calculations in question 14 and what should you do? DO NOT ASSUME ASSIGNABLE CAUSE.
16: The average of a process is 85.02 and the standard deviation is 0.188411.
The specification limits are 85 plus or minus 0.5
a) What is the Cp value?
b) What is the Cpk value?
c) What are your three options for improving your process capability?
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