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Damm [24]
3 years ago
10

The Denver Post reported that, on average, a large shopping center had an incident of shoplifting caught by security 1.4 times e

very eight hours. The shopping center is open from 10 A.M. to 9 P.M. (11 hours). Let r be the number of shoplifting incidents caught by security in an 11-hour period during which the center is open.
(a) Explain why the Poisson probability distribution would be a good choice for the random variable r.
Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are dependent.
Frequency of shoplifting is a rare occurrence. It is reasonable to assume the events are independent.
Frequency of shoplifting is a rare occurrence. It is reasonable to assume the events are dependent.
Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are independent.

What is ? (Round your answer to three decimal places.)


(b) What is the probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security? (Round your answer to four decimal places.)


(c) What is the probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security? (Round your answer to four decimal places.)


(d) What is the probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security? (Round your answer to four decimal places.)
Mathematics
1 answer:
weqwewe [10]3 years ago
6 0

Using the Poisson distribution, it is found that:

  • a) Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are independent.
  • b) 0.8541 = 85.41% probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security.
  • c) 0.303 = 30.3% probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security.
  • d) 0.1459 = 14.59% probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • \mu is the mean in the given interval.

Item a:

In this problem, we are given the mean during an interval, which is a common event(1.4 every 8 hours), and intervals can be assumed to be independent, hence, the correct option is:

  • Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are independent.

Item b:

Mean of 1.4 every 8 hours, hence, every 11 hours, the mean is:

\mu = 11 \times \frac{1.4}{8} = 1.925

The probability is:

P(X \geq 1) = 1 - P(X = 0)

In which:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459

Then

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1459 = 0.8541

0.8541 = 85.41% probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security.

Item c:

The probability is:

P(X \geq 3) = 1 - P(X < 3)

In which:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Then:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459

P(X = 1) = \frac{e^{-1.925}(1.925)^{1}}{(1)!} = 0.2808

P(X = 2) = \frac{e^{-1.925}(1.925)^{2}}{(2)!} = 0.2703

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1459 + 0.2808 + 0.2703 = 0.697

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.697 = 0.303

0.303 = 30.3% probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security.

Item d:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459

0.1459 = 14.59% probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security.

To learn more about the Poisson distribution, you can take a look at brainly.com/question/13971530

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Step-by-step explanation:

The slope-intercept form of a line is y=mx-b where m is slope and b is y-intercept.

The point-slope form of a line is y-y1=m(x-x1) where m is the slope and (x1,y1) is a point on the line.

The standard form a line is ax+by=c.

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So if we are looking for a line parallel to 3x-4y=7 then we need to know the slope of this line so we can find the slope of our parallel line.

3x-4y=7

Goal: Put into slope-intercept form

3x-4y=7

Subtract 3x on both sides:

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Divide both sides by -4:

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Simplify:

   y=\frac{3}{4}x+\frac{-7}{4}

So the slope of this line is 3/4.  So our line that is parallel to this one will have this same slope.

So we know our line should be in the form of y=\frac{3}{4}x+b.

To find b we will use the point that is suppose to be on our new line here which is (x,y)=(-4,-2).

So plugging this in to solve for b now:

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3-2=b

b=1

so the equation of our line in slope-intercept form is y=\frac{3}{4}x+1

So that isn't option 1 because the slope is different.  That was the only option that was in slope-intercept form.

The standard form of a line is ax+by=c and we have 2 options that look like that.

So let's rearrange the line that we just found into that form.

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Clear the fractions because we only want integer coefficients by multiplying both sides by 4.

This gives us:

4y=3x+4

Subtract 3x on both sides:

-3x+4y=4

I don't see this option either.

Multiply both sides by -1:

3x-4y=-4

I do see this as a option. So far the only option that works is 2.

Let's look at point slope form now.

We had the point that our line went through was (x1,y1)=(-4,-2) and the slope,m, was 3/4 (we found this earlier).

y-y1=m(x-x1)

Plug in like so:

y-(-2)=3/4(x-(-4))

y+2=3/4 (x+4)

So option 5 looks good too.

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