Respuesta :
Answer:
0.0049 = 0.49% probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month.
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
n instances of a normal variable:
For n instances of a normal variable, the mean is:
[tex]M = n\mu[/tex]
[tex]s = \sigma\sqrt{n}[/tex]
Mean of 4 minutes, standard deviation of 0.5 minutes:
This means that [tex]\mu = 4, \sigma = \sqrt{0.5}[/tex]
30 days:
[tex]M = 30(4) = 120[/tex]
[tex]s = \sqrt{0.5}\sqrt{30} = \sqrt{0.5*30} = \sqrt{15}[/tex]
What is the probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month (30 days)?
2 hours and 10 minutes is 2*60 + 10 = 130 minutes, so this probability is 1 subtracted by the p-value of Z when X = 130. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this context, due to the 30 instances of the normal variable:
[tex]Z = \frac{X - M}{s}[/tex]
[tex]Z = \frac{130 - 120}{\sqrt{15}}[/tex]
[tex]Z = 2.58[/tex]
[tex]Z = 2.58[/tex] has a p-value of 0.9951.
1 - 0.9951 = 0.0049
0.0049 = 0.49% probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month.
