You are making 20 pound bags of charcoal that people buy for their backyard gril

Question

You are making 20 pound bags of charcoal that people buy for their backyard grills. Since it is hard to fill exactly 20 pounds / bag, you decide to overfill each bag by about half a pound. The standard deviation for your production process is 1 pounds. Answer all of the following questions: 11. If you pick up a bag at random, what is the probability that it will have less than 20 pounds? Mean = 20.5, sigma = 1 12. If you pick up a sample of 12 bags at random, what is the probability that the sample mean weight will be less than 20 pounds? 13. Suppose you pick 15 bags and the average weight of these bags is 20.12 pounds. Construct 99% confidence intervals. What are the LCL and UCL? 14. You Want to estimate the population mean very tightly – within .1 pounds around the sample mean; that is, you want to be 95% confident that the population mean of all the bags will lie between +0.1 and -0.1 pounds of the sample mean. What should be the sample size that you should select?

Explanation / Answer

11.
Mean ( u ) =20.5
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X < 20) = (20-20.5)/1
= -0.5/1= -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.3085                  
12.
Mean ( u ) =20.5
Standard Deviation ( sd )= 1/ Sqrt(n) = 0.2887
Number ( n ) = 12
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X < 20) = (20-20.5)/1/ Sqrt ( 12 )
= -0.5/0.2887= -1.7321
= P ( Z <-1.7321) From Standard NOrmal Table
= 0.0416                  
13.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=20.12
Standard deviation( sd )=1
Sample Size(n)=15
Confidence Interval = [ 20.12 ± Z a/2 ( 1/ Sqrt ( 15) ) ]
= [ 20.12 - 2.58 * (0.26) , 20.12 + 2.58 * (0.26) ]
= [ 19.45,20.79 ]

14.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1
ME =0.1
n = ( 1.96*1/0.1) ^2
= (1.96/0.1 ) ^2
= 384.16 ~ 385      

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.