El teorema de wilson

Mathematics

1 answer:

My name is Ann [436]3 years ago

6 0

Answer:

En matemáticas, especialmente en la teoría de números hay una proposición que vincula tres conceptos: primalidad, factorial de un número entero no nulo y congruencia de números respecto de un módulo.

answer-Wilson's theorem, in number theory, theorem that any prime p divides (p − 1)! ... + 1, where n! is the factorial notation for 1 × 2 × 3 × 4 × ⋯ × n. For example, 5 divides (5 − 1)!

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At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine

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Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

Step-by-step explanation:

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

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 z-score 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 p-value, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$