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Setler79 [48]
3 years ago
9

The times to process orders at the service counter of a pharmacy are exponentially distributed with mean 1 0 minutes. If 100 cus

tomers visit the counter in a 2-day period, what is the probability that at least half of them need to wait more than 10 minutes?
Mathematics
1 answer:
g100num [7]3 years ago
8 0

Answer:

Therefore,  the probability that at least half of them need to wait more than 10 minutes is <em>0.0031</em>.

Step-by-step explanation:

The formula for the probability of an exponential distribution is:

P(x < b) = 1 - e^(b/3)

Using the complement rule, we can determine the probability of a customer having to wait more than 10 minutes, by:

p = P(x > 10)

  = 1 - P(x < 10)

  = 1 - (1 - e^(-10/10) )

  = e⁻¹

  = 0.3679

The z-score is the difference in sample size and the population mean, divided by the standard deviation:

z = (p' - p) / √[p(1 - p) / n]

  = (0.5 - 0.3679) / √[0.3679(1 - 0.3679) / 100)]

  = 2.7393

Therefore, using the probability table, you find that the corresponding probability is:

P(p' ≥ 0.5) = P(z > 2.7393)

<em>P(p' ≥ 0.5) = 0.0031</em>

<em></em>

Therefore,  the probability that at least half of them need to wait more than 10 minutes is <em>0.0031</em>.

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