Respuesta :
Answer:
a) 51.60% probability that in a given year there will be less than 21 earthquakes.
b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.
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 = 20.8, \sigma = 4.5[/tex]
a) Find the probability that in a given year there will be less than 21 earthquakes.
This is the pvalue of Z when X = 21. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21 - 20.8}{4.5}[/tex]
[tex]Z = 0.04[/tex]
[tex]Z = 0.04[/tex] has a pvalue of 0.5160.
So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.
b) Find the probability that in a given year there will be between 18 and 23 earthquakes.
This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:
X = 23
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{23 - 20.8}{4.5}[/tex]
[tex]Z = 0.71[/tex]
[tex]Z = 0.71[/tex] has a pvalue of 0.7611
X = 18
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{18 - 20.8}{4.5}[/tex]
[tex]Z = -0.62[/tex]
[tex]Z = -0.62[/tex] has a pvalue of 0.2676
So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.
