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

User Ryan Grush
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Answer:

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

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)

P(p' ≥ 0.5) = 0.0031

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

User Rolnn
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