Mathematics, 05.03.2021 01:50 dizzy1111

Suppose that, on the average, 1/3 of the graduating seniors at a certain college have two parents attend the graduation ceremony, another third of these seniors have one parent attend the ceremony, and the remaining third of these seniors have no parents attend. If there are 600 graduating seniors in a particular class, what is the probability that not more than 650 parents will attend the graduation ceremony

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Answer from: bckyanne3

0.9803 = 98.03% probability that not more than 650 parents will attend the graduation ceremony.

Step-by-step explanation:

To solve this question, we need to know the Poisson and the normal distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval. The standard deviation is the square root of the mean.

Normal Probability Distribution:

Problems of normal distributions can be 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.

The Poisson distribution can be approximated to the normal with same means, and $\sigma = \sqrt{\mu}$

1/3 of the graduating seniors at a certain college have two parents attend the graduation ceremony, another third of these seniors have one parent attend the ceremony, and the remaining third of these seniors have no parents attend.

There are 600 graduating seniors. This means that the expected number of parents attending the ceremony is given by:

$\mu = 600*(2*\frac{1}{3} + 1*\frac{1}{3} + 0*\frac{1}{3}) = 600(\frac{2}{3} + \frac{1}{3}) = 600(1) = 600$

And

$\sigma = \sqrt{600}$

What is the probability that not more than 650 parents will attend the graduation ceremony?

Using continuity correction, as Poisson is a discrete distribution while the normal is continuous, this is $P(X \leq 650 + 0.5) = P(X \leq 650.5)$ , which is the pvalue of Z when X = 650.5 . So

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

$Z = \frac{650.5 - 600}{\sqrt{600}}$

Z = 2.06

Z = 2.06 has a pvalue of 0.9803.

0.9803 = 98.03% probability that not more than 650 parents will attend the graduation ceremony.

Answer from: Quest

i can't read the graph

step-by-step explanation:

Answer from: Quest

$12. that is my answer.

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