Um what is the answer to this picture right here .

Since 1900, the magnitude of earthquakes that measure 0.1 or higher on the Richter Scale in CA are distributed normally with a m

Gwar [14]

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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:

$\mu = 6.2, \sigma = 0.5$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.1 - 6.2}{0.5}$

$Z = 1.8$

$Z = 1.8$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.1 - 6.2}{0.5}$

$Z = -2.2$

$Z = -2.2$ 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 $n = 10, s = \frac{0.5}{\sqrt{10}} = 0.1581$

This is 1 subtracted by the pvalue of when X = 6.1. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{6.1 - 6.2}{0.1581}$

$Z = -0.63$

$Z = -0.63$ 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

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.22 - 6.2}{0.1581}$

$Z = 6.45$

$Z = 6.45$ has a pvalue of 1

X = 5.7

$Z = \frac{X - \mu}{s}$

$Z = \frac{5.7 - 6.2}{0.1581}$

$Z = -3.16$

$Z = -3.16$ 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.

$Z = \frac{X - \mu}{\sigma}$

$-0.253 = \frac{X - 6.2}{0.5}$

$X - 6.2 = -0.253*0.5$

$X = 6.0735$

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