During late-night hours, the 10^th Street convenience store averages four customers per hour.No customers have arrived in the last half-hour.Assuming that all Poisson conditions are met, determine the probability that the next customer arrives in less than 10 minutes. A).3465 B).4866 C).3882 D).2989 E).4356

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Quizplus · Accepted Answer

Given that the store averages four customers per hour, the arrival rate ($\lambda$) is 4 customers per hour. For a time interval of 10 minutes (which is $\frac{1}{6}$ of an hour), the expected number of arrivals is $\lambda \times \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$. The probability that the next customer arrives in less than 10 minutes can be found using the Poisson distribution formula for 0 arrivals in the given time period and then subtracting this from 1 to find the probability of at least one arrival. The formula for the probability of exactly k arrivals is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$. For 0 arrivals, $P(X=0) = e^{-\lambda}$, where $\lambda = \frac{2}{3}$ for the 10-minute interval. Thus, $P(X \geq 1) = 1 - P(X=0) = 1 - e^{-\frac{2}{3}}$. Calculating $e^{-\frac{2}{3}}$ and subtracting from 1 gives approximately 0.4866, or 48.66%.

During Late-Night Hours, the 10th Street Convenience Store Averages Four

During Late-Night Hours, the 10^th Street Convenience Store Averages Four