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gtnhenbr [62]
2 years ago
14

The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the proba

bility of an occurrence is the same in any two-time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3. The probability that there are 8 occurrences in ten minutes is _________.
Mathematics
1 answer:
Olegator [25]2 years ago
3 0

Answer:

The probability that occurrence of 8 in ten minutes is 0.0771.

Step-by-step explanation:

Poisson Distribution:

A discrete random variable X having the enumerable set {0,1,2,.....} as the spectrum, is said to have Poisson distribution with parameter \mu (>0), if the p.m.f is given by

P(X=x)=\frac{e^{-mu}\mu^x}{x!}  for x=0,1,2,...

               = 0,           elsewhere

The mean number of occurrences in ten minutes is 5.3.

Here \mu = 5.3 and x= 8

P(X=8)=\frac{e^{-5.3}(5.3)^8}{8!}

               =0.0771

The probability that occurrence of 8 in ten minutes is 0.0771.

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A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
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Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

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Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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