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Anika [276]
3 years ago
14

The Department of Commerce in a particular state has determined that the number of small businesses that declare bankruptcy per

month is approximately a Poisson distribution with a mean of 6.4. Find the probability that exactly 5 bankruptcies occur next month.
A) 0.0589
B) 0.1487
C) 0.1987
D) 0.2987
Mathematics
1 answer:
skad [1K]3 years ago
4 0

Answer:

B) 0.1487

Step-by-step explanation:

Let X be the discrete random variable that represents the number of events observed over a given time period.  If X follows a Poisson distribution, then the probability of observing k events over the time period is:

P(X=k)=\frac{\lambda^{k} *e^{-\lambda} }{k!}

Where:

\lambda=Mean\\k=number\hspace{3}of\hspace{3}events\\e=Euler's\hspace{3}number

So, the probability that exactly 5 bankruptcies occur next month is:

P(X=5)=\frac{6.4^{5} *e^{-6.4} }{5!} =\frac{17.84083537}{120} =0.1486736281\approx0.1487

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Think of a number, divide it by 5 and then add 27 to the result. The resulting number is half of the original number. What was t
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100 = my number
5 = divide by

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27 = add
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A researcher obtains t = 2.35 for a repeated-measures study using a sample of n = 8 participants. Based on this t value, what is
ZanzabumX [31]

Answer:

p_v =2*P(t_{7}>2.35)=2*0.0255=0.051  

If we compare the p value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% or 1% of significance we fail to reject the null hypothesis.

Step-by-step explanation:

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level is not provided but we can assume it as \alpha=0.05. First we need to calculate the degrees of freedom like this:

df=n-1=8-1=7

The next step would be calculate the p value for this test.  Since is a bilateral test or two tailed test, the p value would be:  

p_v =2*P(t_{7}>2.35)=2*0.0255=0.051  

If we compare the p value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% or 1% of significance we fail to reject the null hypothesis.  

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3 years ago
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