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Anna71 [15]
3 years ago
6

Customers arrive randomly at a service point at an average rate of 30 per hours.Assuming a Possition distibution,calculate the p

robability that
a.no customers arrive in a particular minute.
b.exactly one customer arrives in five minutes.
c.at most one customer arrive in ten minutes.​
Mathematics
1 answer:
inna [77]3 years ago
4 0

Answer:

honestly man I say go with c

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n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

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".  

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Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.05 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We can use as an estimator for p \hat p =0.5. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

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