Respuesta :
Answer:
Method A.
Step-by-step explanation:
For solving this question we need to find out the z-scores for both methods,
Since, the z-score formula is,
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where, [tex]\mu[/tex] is mean,
[tex]\sigma[/tex] is standard deviation,
Given,
For method A,
[tex]\mu = 33[/tex]
[tex]\sigma=8[/tex]
Thus, the z score for 34 is,
[tex]z_1=\frac{34-33}{8}=0.125[/tex]
While, for method B,
[tex]\mu = 37[/tex]
[tex]\sigma = 4[/tex]
Thus, the z score for 34 is,
[tex]z_2=\frac{34-37}{4}=-0.75[/tex],
Since, [tex]z_1 > z_2[/tex]
Hence, method A is preferred if the procedure must be completed within 34 minutes.
Comparison of two normal distribution can be done via intermediary standard normal distribution. The procedure to be preferred for getting the procedure completed within 34 minutes is: Method A
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
Firstly, we need to figure out what the problem is asking, and a method which we can apply. Two normal distributions have to be compared. We can convert them to standard normal distribution for comparison. Then, we will get the p-value for 34 minutes(converted to standard normal variate's value) which will tell about the probability of obtaining time as 34 minutes(or under it)(this can be obtained with p value) in method A or B. The more the probability is there, the more chances for that method would be for getting completed within 34 minutes (compared to other method).
- For method A:
Let X = time taken for completion of surgical procedure by method A,
Then, by given data, we have: [tex]X \sim N(\mu = 33, \sigma = 8)[/tex]
The probability that X will fall within value 34 is [tex]P(X \leq 34)[/tex]
Converting this whole thing to standard normal distribution, we get the needed probability as:
[tex]P(X \leq 34) = P(Z = \dfrac{X - \mu}{\sigma} \leq \dfrac{34 - 33}{8} ) = P(Z \leq 0.15)[/tex]
From the z-tables, the p value for Z = 0.15 is 0.5596
Thus, we get:
[tex]P(X \leq 34) = P(Z \leq 0.15 ) \approx 0.5596[/tex]
- For method B:
Let Y = time taken for completion of surgical procedure by method B,
Then, by given data, we have: [tex]Y \sim N(\mu = 37, \sigma = 4)[/tex]
The probability that X will fall within value 34 is [tex]P(Y \leq 34)[/tex]
Converting this whole thing to standard normal distribution, we get the needed probability as:
[tex]P(Y \leq 34) = P(Z = \dfrac{Y - \mu}{\sigma} \leq \dfrac{34 - 37}{4} ) = P(Z \leq -0.25)[/tex]
From the z-tables, the p value for Z = -0.25 is 0.4013
Thus, we get:
[tex]P(Y \leq 34) = P(Z \leq -0.25 ) \approx 0.4013[/tex]
Thus, we see that:
P(method A will make surgical procedure last within 34 minutes) = 0.5596 > P(method B will make surgical procedure last within 34 minutes) = 0.4013
Thus, method A should be preferred, as there is higher chances for method A to get the surgery completed within 34 minutes than method B.
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
