A Company produces candy on a production line. When operating properly, the production process will produce an average of u=140 pounds of candy per hour with a standard deviation of o=20 pounds of candy per hour. It is also known that the distribution is skewed left. In order to monitor the assembly process, 100 hours are randomly sampled and the number of pounds of candy produced is recorded. The sample mean number produced x-bar will be used to determine whether the production operation is working correctly.

-Can you explain to me how the sampling distribution of x-bar is generated theoretically and what would be the property(ies) of this x-bar distribution when the production is working properly?
-If the sample mean of the x-bar is 70 for the 100 hours...does this mean that the assembly is producing too few pounds of candy on average?

Thank you very much!

Solution Summary

This solution discusses X-bar distribution and sampling in over 200 words.

... Expected shape is normal distribution as sample size is ... µ= Standard deviation =σ= $40,000 50 sample size=n ... σx=standard error of mean=σ/√n= xbar= $112,000 z ...

...Distribution. Probability for sample means 64 Mean=µ= Standard deviation =σ= 2 4 sample size=n= = ( 2 /√ 4) σx=standard error of mean=σ/√n= 1 xbar= 66 z ...

... is Prob(Z)= 0.8413 Therefore probability corresponding to xbar> 94.00 is 1 ... the mean and the standard deviation for the sampling distribution of x-bar for n ...

... 3) and the shape of the sampling distribution of the mean approaches normal as the ... root of n= 0.05 = ( 0.4 /square root of 64) sample mean= xbar= 3 minutes ...

... population variance is 10.16, P(-1<xbar-mean<1 ... and draw a dotplot for sampling distribution of the ... to create one plot with the sampling distributions for each ...

Sampling distribution: Sample Means; Bike shop mean repair cost exceeding $16. ... the probability the mean repair cost for a sample of 100 ... (a) z = (x-bar - μ)/(σ ...

... σ / n = 100 / 75 = 11.55 A. What is the probability that a random sample of 75 students will provide a sample mean SAT ...xbar − µ P(1010<=X bar<=1030)=P ...

... the distribution of amounts purchased follows the no distribution. ... Mean=µ= Standard deviation =σ= $5.00 50 sample size=n ... of mean=σ/√n= 0.7071 xbar= $25.00 z ...

... corresponding to xbar> 3.34 is 1-Prob(Z)= 0.0025 =1-0.9975 0r= 0.25%. Answer: 0.25%. Calculates probabilities for sample means using Normal Distribution. ...

... We will use normal distribution as sample size (50) is large and population ... root of n= 0.02121 = ( 0.15 /square root of 50) sample mean=xbar= 0.99 liters z ...