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ahrayia [7]
3 years ago
15

The waiting time, in minutes, to see a teller at a large bank follows an exponential distribution. If the proportion of all cust

omers who wait more than 15 minutes is 0.01, what is the mean waiting time for all customers?
Mathematics
1 answer:
Arada [10]3 years ago
4 0

Answer: 0.00067 minutes

Step-by-step explanation: if the proportion of customers who wait more than 15 minutes is 0.01, then the time interval between each waiting customer 15/0.01 = 1500 minutes.

The distribution that defines this question is that of an exponential.

An exponential distribution is dependent on the fixed time rate at which the event is occurring (λ)

For this question of ours, λ = 1500 minutes.

The mean of an exponential distribution is given as

u = 1/ λ = 1/1500 = 0.00067 minutes.

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Answer:

a) \bar X \sim N(\mu=375, \sigma={\bar X}=\frac{7}{\sqrt{40}}=1.107)

b) Since the sample size is large enough n>30 and the original distribution for the random variable X  doesn’t appear to have extreme skewness or outliers, the distribution for the sample mean would be bell shaped and symmetrical.

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Previous concepts

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The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable of interest. We know from the problem that the distribution for the random variable X is given by:

E(X) = 75

sd(X) = 7

We take a sample of n=40 . That represent the sample size

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

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Part b

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In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X \leq 77)=P(Z

We can us the following excel code: "=NORM.DIST(1.807,0,1,TRUE)"

Part d

See the figure attached.

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