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galina1969 [7]
3 years ago
9

If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of waiting more than 0.5 hour for the next

arrival is:
Mathematics
1 answer:
solniwko [45]3 years ago
6 0
<span>The answer is 0.8347. Mean u = 1/m = 3.6 per hour Time x = 0.5 hour Exponential Probability = P(X <= x) Going by the Cumulative distribution function P(X <= x) = 1 - e^(-mx) P(X < .5) = 1 – exp^(-3.6 x 0.5) = 1 – 0.1653 = 0.8347</span>
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Answer:

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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 100, \sigma = 5

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This is 1 subtracted by the pvalue of Z when X = 110. So

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Z = 2

Z = 2 has a pvalue of 0.9772

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