ISYE 3125 - HW2_Ch03DONE
pdf
keyboard_arrow_up
School
Kennesaw State University *
*We aren’t endorsed by this school
Course
4543
Subject
Industrial Engineering
Date
Feb 20, 2024
Type
Pages
4
Uploaded by GrandEchidnaPerson1070
1 Name: Michael Williams ISYE 3125 – Statistical Quality Control Ch03 – Modeling Process Quality HW2 This HW consists of five questions and is worth 100 points. Please show all work
for problems requiring computations. Posting only the results earns zero points. Problem 3.1 The content of liquid detergent bottles is being analyzed. Twelve bottles, randomly selected from the process, are measured, and the results are as follows (in fluid ounces): 16.05, 16.03, 16.02, 16.04, 16.05, 16.01, 16.02, 16.02, 16.03, 16.01, 16.00, 16.07. A.
Calculate the sample average. (4 pts) (16.05 + 16.03 + 16.02 + 16.04 + 16.05 + 16.01 + 16.02 + 16.02 + 16.03 + 16.01+ 16.00 +16.07)/ 12 = 16.029 B. Calculate the sample standard deviation. (
4 pts) Sample mean = 16.0292 Squared differences:
(16.05
−
16.02916667) ^2
(16.03
−
16.02916667) ^2 (16.02
−
16.02916667) ^2 (16.04
−
16.02916667) ^2 (16.05
−
16.02916667) ^2 (16.01−16.
02916667) ^2 (16.02−16.
02916667) ^2 (16.02−16.
02916667) ^2 (16.03−16.
02916667) ^2 (16.01−16.
02916667) ^2 (16.00
−
16.02916667) ^2 (16.07−16.
02916667) ^2 Variance is SUM OF Squared differences DIVIDED BY 11
Sum of Squared = 0.004651 0.005406 / 11 = 0.0004228 which is the variance
Sample Standard Deviation is the Square root of the variance so √
0.0004505 = 0.020 is Sample Standard deviation
2 Problem 3.4 Mean=6+26+8+2+6+3+10+14+4+5+3+17+9+8+9+5+3+28+21+4 / 20 = 9.55 I will use (…) for most of the squared differences
to find the variance because there are 20 measurements and It makes things easier to read but I am still showing my work and process
3 Problem 3.17 The normal probability plot seems like the best one to use because it is visually easier to interpret and understand where a of the concentration is on the plot. -1
0
1
2
3
4
5
6
0
20
40
60
80
100
120
140
Z Value
Concentration of Suspended Solid (ppm)
Normal Probability Plot
0
0.005
0.01
0.015
0.02
0.025
-1
0
1
2
3
4
5
6
pdf
x
Log-Normal Plot
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
4 Problem 3.25 Construct a box plot for the data in Exercise 3.1 and interpret the elements of the box plot. (4 pts) There are 4 outliers and the data is skewed left just a bit. Since the mean is less than the median.