Respuesta :
Answer:
a) 0.5447
b) 0.0228
c) 0.4325
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.
Normal distribution with a mean of 9.6 minutes and a standard deviation of 2.3 minutes.
This means that [tex]\mu = 9.6, \sigma = 2.3[/tex]
a. between 5 and 10 min
This is the p-value of Z when X = 10 subtracted by the p-value of Z when X = 5. So
X = 10
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10 - 9.6}{2.3}[/tex]
[tex]Z = 0.17[/tex]
[tex]Z = 0.17[/tex] has a p-value of 0.5675
X = 5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 9.6}{2.3}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a p-value of 0.0228
0.5675 - 0.0228 = 0.5447 probability that a randomly received emergency call is between 5 and 10 minutes.
b. less than 5 min
p-value of Z when X = 5, which from item a), is 0.0228, so 0.0228 probability that a randomly received emergency call is of less than 5 minutes.
c. more than 10 min
1 subtracted by the p-value of Z when X = 10, which, from item a), is of 0.5675.
1 - 0.5675 = 0.4325
0.4325 probability that a randomly received emergency call is of more than 10 minutes.
