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2 years ago

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Mathematics

1 answer:

vova2212 [387]2 years ago

3 0

Answer:

n = 6

n, = 6n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2

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A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc

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Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

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

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .