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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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Step-by-step explanation:

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Raji has 5/7 as many CDs as Megan. If Raji gives 1/10 of her CDs to Megan, what will be the ratio of the number of Raji’s CDs to
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3 years ago
The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

Z = \frac{X - \mu}{\sigma}

By the Central limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
How do u graph linear inequality
amm1812
There are three steps

1 rearrange the equation so "y" is on the left and everything else is on the right

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3 shade above the line for a "greater than" (y> or y>) line under that or below the line for a less than (y< or y< line under it)
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Help me please! so jarring
pishuonlain [190]

Answer:

y = -5x + 4.

Step-by-step explanation:

y = mx + c  is the general form

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The graph passes through the points (0, 4) and (1, -1) so the slope

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= (-1-4) / (1-0)

= -5 = m.

and c = 4.

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2 years ago
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