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.
If using the empirical rule, since 13.9 hits is more than two standard deviations above the mean of 10.2 hits, the process could be considered out of control. However, establishing specific control limits is necessary for a definitive answer.
Explanation:To determine whether a process is in control or out of control, we assess whether a given measurement is within the expected range of a normal distribution, often using the empirical rule or control limits. Given that the process has a mean of 10.2 hits per minute and a standard deviation of 1.04 hits, under the empirical rule, approximately 95% of the data should fall within two standard deviations of the mean (that is, between roughly 8.12 and 12.28 hits).
With 13.9 hits in a randomly selected minute, this count is significantly more than two standard deviations above the mean, suggesting that the process might be out of control. However, to make a definitive statement about control status, specific control limits must be established, often based on the particular specifications of the process being monitored.
An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement. find the probability of drawing 2 blue and 1 red given at least 1 blue is drawn
Answer:
Frist case: P=12/35
Second case: P=31/35
Step-by-step explanation:
An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement.
Frist case:
We calculate the number of possible combinations
{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35
We calculate the number of favorable combinations
{3}_C_{2} · {4}_C_{1} =
=\frac{3!}{2! · (3-2)!} · \frac{4!}{1! · (4-1)!}
=3 · 4 = 12
Therefore, the probability is
P=12/35
Second case:
When we count on at least one ball to be blue, we go over the probability complement.
We calculate the probability that all the balls are red, then subtract this from 1.
We calculate the number of possible combinations
{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35
We calculate the number of favorable combinations
{4}_C_{3} = \frac{4!}{3! · (4-3)!=4
The probability is
P=4/35.
Therefore the probability on at least one ball to be blue
P=1-4/35
P=31/35