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Aleksandr-060686 [28]
3 years ago
12

The number of clients arriving at a bank machine is Poisson distributed with an average of 2 per minute. What is the probability

that no more than 2 customers will arrive in a minute
Mathematics
1 answer:
IrinaVladis [17]3 years ago
6 0

Answer:

The probability that no more than '2' customers will arrive in minute = 1.5412

Step-by-step explanation:

Explanation:-

Mean of the Poisson distribution 'λ' = 2 per minute

P(X=x) = e⁻ˣ λˣ/x!

The probability that no more than '2' customers will arrive in minute

P(x≤ 2) = P(x=0) +P(x=1)+P(x=2)

           = e⁻² (2)°/0! + e⁻²(2)¹/1!+e⁻² (2)²/2!

          = 1 + 0.2706 + 0.2706

          = 1.5412

The probability that no more than '2' customers will arrive in minute = 1.5412

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The distribution of IQ scores can be modeled by a normal distribution with mean 100 and standard deviation 15.
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Answer:

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Step-by-step explanation:

We are given the following information in the question:

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We are given that the distribution of IQ scores is a bell shaped distribution that is a normal distribution.

a) Let X be a person's IQ score.

Then, density functions for IQ scores is given by:

P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{z^2}{2}}\\\\\text{where,}\\\\z = \frac{x-\mu}{\sigma}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\\\\P(x) = \displaystyle\frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-100)^2}{450}}

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Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(120 \leq x \leq 125) = P(\displaystyle\frac{120 - 100}{15} \leq z \leq \displaystyle\frac{125-100}{15}) = P(1.33 \leq z \leq 1.66)\\\\= P(z \leq 1.66) - P(z < 1.33)\\= 0.952 - 0.908 = 0.044 = 4.4\%

P(120 \leq x \leq 125) = 4.4\%

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