Respuesta :
Answer:
This process is out of control.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
A probability is said to be unusual if it's z-score has a pvalue of 0.05 or lower, or a pvalue of 0.95 or higher.
In this problem, we have that:
[tex]\mu = 10.2 \sigma = 1.04[/tex]
If a randomly selected minute has 13.9 hits, would the process be considered in control or out of control?
The process will be considered out of control if it's z-score(Z when X = 13.9) has a pvalue of 0.95 or higher. Otherwise(it will be positive, since 13.9 is above the mean), it will be considered in control.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{13.9 - 10.2}{1.04}[/tex]
[tex]Z = 3.56[/tex]
[tex]Z = 3.56[/tex] has a pvalue of 0.9999. So there is only a 1-0.9999 = 0.0001 = 0.01% probability of getting 13.9 hits a minute.
So this process is out of control.
