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3 years ago

PLEASE HELP

1 answer:

8 0

If an event occurs 0 times (out of 4, in this case) then it does not occur at least once. So we can find the probability of it not occurring and then subtract that value from 1.

So, what are the chances of it not occurring in 1 trip?

1−.37=.63

What about not occurring in 2 trips?

(1−.37)⋅(1−.37)=.63⋅.63=.3969

Now what about not occurring in 4 trips?

.63^4 = 0.15752961

We must subtract this value from 1

(recall that what we just calculated is the probability of it not occurring, so the probability of it occurring at least once is:

1−0.15752961 = .84247039

TLDR - In 4 trips the chance of a guest catching a cutthroat once in under 4 trips in 0.84.

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Read 2 more answers

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

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Answer:

a)From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

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3 years ago

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Step-by-step explanation:

a solution is a system of equations intersecting.

on the graph you can see that the lines intersect at (-2,2)

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2 years ago