: Fifty soft drink bottles of a specific brand are collected from one day produc
ID: 3202997 • Letter: #
Question
: Fifty soft drink bottles of a specific brand are collected from one day production and measured its net weight, which are given below: The specification limits for this brand are (16+0.5oz)
15.8 16.3 16.2 16.1 16.6
16.3 15.9 15.9 16.2 16.4
16.1 16.2 16.5 16.4 16.3
16.3 16.2 15.9 16.4 16.2
16.1 16.1 16.4 16.5 16.0
16.1 15.8 16.7 16.6 16.4
16.1 16.3 16.5 16.1 16.5
16.2 16.1 16.2 16.1 16.3
16.3 16.2 16.4 16.3 16.5
16.6 16.3 16.4 16.1 16.5
1. Estimate the mean and standard deviation
2. Draw a histogram with superimposing the specification limits. Interpret the histogram focussing on how to improve the process.
3. Draw normal probability plot to justify your answer in (2)
4. Assuming the normality, estimate the sigma level of the process
5. Estimate the percentage of soft drink bottle out side the lower specification and upper specification.
6. Since the soft drink bottles fell outside the upper specification are quite large, it is decided to lower the process mean setting by 0.2 units. Also note that soft drink bottles with net weight less than the lower specification are usually rejected where as bottles with net weight more than upper specification will not be rejected. If you are the production manager, how many bottles of drink you need to produce to get 10000 accepted bottles, with new process mean setting, but same standard deviation.
Use R studio and give out proper solutions.
Explanation / Answer
R code & answers:
bottle=c(15.8, 16.3, 16.2, 16.1, 16.6, 16.3, 15.9, 15.9, 16.2, 16.4,16.1, 16.2, 16.5, 16.4, 16.3, 16.3, 16.2, 15.9, 16.4, 16.2, 16.1, 16.1, 16.4, 16.5, 16.0,16.1, 15.8, 16.7, 16.6, 16.4, 16.1, 16.3, 16.5, 16.1, 16.5,16.2, 16.1, 16.2, 16.1, 16.3,16.3, 16.2, 16.4, 16.3, 16.5,16.6, 16.3, 16.4, 16.1, 16.5)
#average is 16.258
mean(bottle)
#1. STD DEV:0.212
Std_Dev=sd(bottle)
#2. Draw a histogram with superimposing the specification limits. Interpret the histogram focussing on how to improve the process
hist(bottle)
#interpretation: distribution is close to Normal distribution but positively skewed.
#control limits: 15.5 & 16.5
#3. Draw normal probability plot
qqnorm(bottle)
#from Q-Q normal plot we can say that: distribution is close to Normal distribution but positively skewed.
#4. control limits: 15.5 & 16.5
#5 (using excel):
P(X<15.5) 0.917% =NORM.DIST(15.5,16,0.212,TRUE) P(X>16.5) 0.917% =1-NORM.DIST(16.5,16,0.212,TRUE)Related Questions
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