Use properties to find the sum or product of 93+(68+7)

What is the probability that a randomly selected member of a normally distributed population will lie more than 1.8 standard dev

Helen [10]

Answer:

$P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)$

And we can find this probability using the complement rule:

$P(z>1.8)=1-P(z$

And using the normal standard distirbution table or excel we got:

$P(z>1.8)=1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable if interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sigma)$

We are interested on this probability

$P(X>\mu +1.8 \sigma)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X> \mu +1.8\sigma)=P(\frac{X-\mu}{\sigma}>\frac{\mu +1.8\sigma-\mu}{\sigma})=P(Z>1.8)$

And we can find this probability using the complement rule:

$P(z>1.8)=1-P(z$

And using the normal standard distirbution table or excel we got:

$P(z>1.8)=1-P(z$

5 0

3 years ago

6m+2÷8=37 what does m=

Mama L [17]

6m+2/=37×8
6m=296-2
6m=294
m=49

4 0

3 years ago

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If F(x) = 4%-3, which of the following is the inverse of F(x)?

Brums [2.3K]

I assume you meant to type x for %.

If not, basically back-substitute at the end of the solution.

F(x) = 4x - 3

Let y = F(x)

y = 4x - 3

x = 4y - 3

x - 3 = 4y

(x - 3)/4 = y

Let y = F^(-1)x

F^(-1)x = (x - 3)/4

If you want % in your answer, just replace x for % in your answer.

8 0

3 years ago

Which statement best describes the meaning of the term average waiting-<br> time?

kotegsom [21]

Answer:

The average number of trials required to get the first success

5 0

3 years ago

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir

Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, we find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

$\sigma = \sqrt{484} = 22$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$5 = 1.96*\frac{22}{\sqrt{n}}$

$5\sqrt{n} = 43.12$

$\sqrt{n} = \frac{43.12}{5}$

$\sqrt{n} = 8.624$

$(\sqrt{n})^{2} = (8.624)^{2}$

$n = 74.4$

We need a sample size of at least 75.

6 0

3 years ago

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