Respuesta :
Answer:
22.96% probability that a hotel room costs between $250.00 and $285.00
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.
In this problem, we have that:
[tex]\mu = 244, \sigma = 55[/tex]
What is the probability that a hotel room costs between $250.00 and $285.00?
This is the pvalue of Z when X = 285 subtracted by the pvalue of Z when X = 250. So
X = 285
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{285 - 244}{55}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a pvalue of 0.7734
X = 250
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{250 - 244}{55}[/tex]
[tex]Z = 0.11[/tex]
[tex]Z = 0.11[/tex] has a pvalue of 0.5438
0.7734 - 0.5438 = 0.2296
22.96% probability that a hotel room costs between $250.00 and $285.00
