Respuesta :
Answer:
Let X the random variable that represent the time to service a customer complaint of a population, and for this case we know the following info
Where [tex]\mu=3[/tex] and [tex]\sigma=0.5[/tex]
And we want to estimat ethe lower control level for with a sample size of n=10 and using the 3 sigma rule. For this case the sample mean is assumed normally distributed.
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And the formula for the lower control limit would be:
[tex] \bar X -3 \frac{\sigma}{s\qrt{10}}[/tex]
And replacing we got:
[tex] 3 -3 \frac{0.5}{\sqrt{10}}= 2.526[/tex]
So then we can conclude that the best option is:
D) greater than 2.50.
Step-by-step explanation:
The LCL for the chart would be:
A) less than 2.40.
B) greater than 2.40 but less than or equal to 2.45.
C) greater than 2.45 but less than or equal to 2.50.
D) greater than 2.50.
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the time to service a customer complaint of a population, and for this case we know the following info
Where [tex]\mu=3[/tex] and [tex]\sigma=0.5[/tex]
And we want to estimat ethe lower control level for with a sample size of n=10 and using the 3 sigma rule. For this case the sample mean is assumed normally distributed:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And the formula for the lower control limit would be:
[tex] \bar X -3 \frac{\sigma}{s\qrt{10}}[/tex]
And replacing we got:
[tex] 3 -3 \frac{0.5}{\sqrt{10}}= 2.526[/tex]
So then we can conclude that the best option is:
D) greater than 2.50.
