In: Statistics and Probability
A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.
(d) Predict with 95% confidence the value of the 11th filled container.
(e) Predict with 95% confidence the interval containing 90% of the filled containers from the process.
Since data is normally distributed We have obtained the answer by using Normal distribution using R.
d) To predict the 11th Filled Container paste below mentioned code in R
x=c(12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02,
12.05, 11.99)
# To obtain the Mean
mu= mean(x)
mu
#To obtain the Standard deviation
sd= sd(x)
sd
n=10
#To Predict the Value of 11th Filled Container having 95 % confidence Interval
#95% CI means to be consider as Alpha=0.95
err= qnorm(0.95)*(sd/(sqrt(n)))
# As per Statistic of Normal distribution Sample mean is
unbiased estimator of Population mean
#hence need to obtain the Sample mean i.e. predict11= err+mu
#Question.d
predict11= err+mu
predict11
Ans: Predicted content of 11 th Container: 12.03075
e) To Predict with 95% confidence the interval paste below mentioned code in R
#95% confidence the interval containing 90% of the filled
containers from the process.
# here we need to consider the probality of error as Alpha/2 i.e.
=0.975
err1= qnorm(0.975)*(sd/(sqrt(n)))
# As per Formula to derive C.I of Mean of Normal Distribution
i.e. Lower_Bound= mu- err1
#and Upper_Bound= mu+ err1
lower_CI= mu- err1
lower_CI
upper_CI= mu+ err1
upper_CI
Ans: C.I.= (11.99623, 12.03377 )