Mathematics, 12.03.2020 03:23 zamorano75341

A certain college would like to have 1050 freshmen. This college cannot accommodate more than 1060. Assume that each applicant accepts with probability .6 and that the acceptances can be modeled by a Binomial distribution. If the college takes 1700 students, what is the probability that the college will have too many acceptances (that is, more than 1060)

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Answer from: hgg44137

2.39% probability that the college will have too many acceptances

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

For each item selected, there are only two possible outcomes. Either it is defective, or it is not. This means that we use concepts of the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

n = 1700, p = 0.6

$\mu = E(X) = np = 1700*0.6 = 1020$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1700*0.6*0.4} = 20.2$

If the college takes 1700 students, what is the probability that the college will have too many acceptances (that is, more than 1060)

This is 1 subtracted by the pvalue of Z when X = 1060. So

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

$Z = \frac{1060 - 1020}{20.2}$

Z = 1.98

Z = 1.98 has a pvalue of 0.9761

1 - 0.9761 = 0.0239

2.39% probability that the college will have too many acceptances

Answer from: Quest

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

Answer the following and be the brainiest the mistake in the pattern is that : a. the decimal point

Answer from: Quest

step-by-step explanation:

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