Respuesta :
Answer:
a) 3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1
b) 1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1
c) 73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1
d) 99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22
e) 6.0735
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are 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.
Central Limit Theorem:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 6.2, \sigma = 0.5[/tex]
a.) What is the probability that a randomly selected earthquake in CA has a magnitude greater than 7.1?
This is 1 subtracted by the pvalue of Z when X = 7.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{7.1 - 6.2}{0.5}[/tex]
[tex]Z = 1.8[/tex]
[tex]Z = 1.8[/tex] has a pvalue of 0.9641
1 - 0.9641 = 0.0359
3.59% probability that a randomly selected earthquake in CA has a magnitude greater than 7.1
b.) What is the probability that a randomly selected earthquake in CA has a magnitude less than 5.1?
This is the pvalue of Z when X = 5.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.1 - 6.2}{0.5}[/tex]
[tex]Z = -2.2[/tex]
[tex]Z = -2.2[/tex] has a pvalue of 0.0139
1.39% probability that a randomly selected earthquake in CA has a magnitude less than 5.1
c.) What is the probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1?
Now [tex]n = 10, s = \frac{0.5}{\sqrt{10}} = 0.1581[/tex]
This is 1 subtracted by the pvalue of when X = 6.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{6.1 - 6.2}{0.1581}[/tex]
[tex]Z = -0.63[/tex]
[tex]Z = -0.63[/tex] has a pvalue of 0.2643
1 - 0.2643 = 0.7357
73.57% probability that ten randomly selected earthquakes in CA have mean magnitude greater than 6.1
d.) What is the probability that a ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22
This is the pvalue of Z when X = 7.22 subtracted by the pvalue of Z when X = 5.7. So
X = 7.22
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{7.22 - 6.2}{0.1581}[/tex]
[tex]Z = 6.45[/tex]
[tex]Z = 6.45[/tex] has a pvalue of 1
X = 5.7
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{5.7 - 6.2}{0.1581}[/tex]
[tex]Z = -3.16[/tex]
[tex]Z = -3.16[/tex] has a pvalue of 0.0008
1 - 0.0008 = 0.9992
99.92% probability that ten randomly selected earthquakes in CA have mean magnitude between 5.7 and 7.22
e.) Determine the 40th percentile of the magnitude of earthquakes in CA.
This is X when Z has a pvalue of 0.4. So it is X when Z = -0.253.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.253 = \frac{X - 6.2}{0.5}[/tex]
[tex]X - 6.2 = -0.253*0.5[/tex]
[tex]X = 6.0735[/tex]
