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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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Answer:

(a) Point estimate = 7.10

(b) The critical value is 1.960

(c) Margin of error = 0.800

(d) Confidence Interval = (6.3, 7.9)

(e) We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

Step-by-step explanation:

Given

\bar x = 7.10 -- sample mean

\sigma=5 --- sample standard deviation

n = 150 --- samples

Solving (a): The point estimate

The sample mean can be used as the point estimate.

Hence, the point estimate is 7.10

Solving (b): The critical value

We have:

CI = 90\% --- the confidence interval

Calculate the \alpha level

\alpha = 1 - CI

\alpha = 1 - 90\%

\alpha = 1 - 0.90

\alpha = 0.10

Divide by 2

\frac{\alpha}{2} = 0.10/2

\frac{\alpha}{2} = 0.05

Subtract from 1

1 - \frac{\alpha}{2} = 1 - 0.05

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

From the z table. the critical value for 1 - \frac{\alpha}{2} = 0.95 is:

z = 1.960

Solving (c): Margin of error

This is calculated as:

E = z * \frac{\sigma}{\sqrt n}

E = 1.960 * \frac{5}{\sqrt {150}}

E = 1.960 * \frac{5}{12.25}

E =  \frac{1.960 *5}{12.25}

E =  \frac{9.80}{12.25}

E =  0.800

Solving (d): The confidence interval

This is calculated as:

CI = (\bar x - E, \bar x + E)

CI = (7.10 - 0.800, 7.10 + 0.800)

CI = (6.3, 7.9)

Solving (d): The conclusion

We are 90% confident that the average number of hours worked by the students is between 6.3 and 7.9

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