Respuesta :
Answer:
The company should have placed that guarantee time on 23.8 minutes.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions 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.
Their research showed that these times were normally distributed with a mean time of 17.8 minutes with a standard deviation of 3.2 minutes.
This means that [tex]\mu = 17.8, \sigma = 3.2[/tex]
They wanted to place the guarantee at a time so that only 3% of customers would be expected to wait beyond the guaranteed time.
So the 100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So X when Z = 1.88.
Where should the company have placed that guarantee time
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.88 = \frac{X - 17.8}{3.2}[/tex]
[tex]X - 17.8 = 1.88*3.2[/tex]
[tex]X = 23.8[/tex]
The company should have placed that guarantee time on 23.8 minutes.
