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dusya [7]
3 years ago
15

Considere un proceso industrial en el que se producen vigas de acero las cuales pueden tener defectos de dureza y peso. En el ca

so de un defecto en la dureza requiere de un estudio de laboratorio. Sin embargo, de procesos anteriores se sabe que 12% de las vigas no pasan la prueba de dureza, que 7% no pasan la prueba de pesado y que sólo 0.5% no pasan ninguna de las dos pruebas.
Si en el proceso se elige una viga al azar y una medición rápida identifica que no pasa la prueba de dureza, ¿cuál es la probabilidad de que el pesado no sea el correcto?
Mathematics
1 answer:
ollegr [7]3 years ago
3 0

Step-by-step explanation:

PRUEBA DE HIPÓTESIS” CONTROL ESTADÍSTICO DE PROCESOS “PRUEBA DE HIPÓTESIS” IV – “A”

2. 2 “PRUEBA DE HIPÓTESIS” "AÑO DE LA PROMOCIÓN DE LA INDUSTRIA RESPONSABLE Y DEL COMPROMISO CLIMÁTICO" UNIVERSIDAD NACIONAL “SAN LUIS GONZAGA” DE ICA FACULTAD DE ADMINISTRACIÓN “PRUEBA DE HIPÓTESIS” Curso : CONTROL ESTADÍSTICO DE PROCESOS Doctor : Dr. ORLANDO GABRIEL HERNÁNDEZ Año : IV - “A” Turno : MAÑANA Autores : CARRILLO MAMANI, GERALD FLORES JAUJE, LUIS HUARIPAUCAR GAMBOA, SONIA OCHOA SUAREZ, GIUSEPI 2014

3. 3 “PRUEBA DE HIPÓTESIS” INTRODUCCIÓN Es evidente que las distribuciones muestrales, vistas en el capítulo anterior, basadas en la teoría de la distribución normal, desarrollan un papel de gran importancia en la inferencia estadística. La inferencia estadística comprende dos partes principales, a saber: la estimación de parámetros y la prueba o docimasia de hipótesis. En este capítulo estudiaremos la segunda de ellas, con el fin de desarrollar métodos y observar su aplicación a problemas corrientes de la vida diaria. La inferencia estadística está basada en el supuesto de tomar muchas muestras, todas con igual probabilidad de ser seleccionadas y a través de una de ellas sabremos algo acerca de la población, mediante el cálculo de estimadores, que nos permitan hacer aseveraciones, incorrectas algunas veces, estableciéndose la probabilidad de error. Este método se basa en la aplicación de técnicas de muestreo, para lo cual se requiere de un buen diseño, además de la aplicación de métodos aleatorios de selección, cuando las probabilidades son iguales para cada elemento de una población. En algunos casos no requieren ser iguales, siempre que se conozcan y sean diferentes a cero. CONTENIDO  Conceptos generales, usos y procedimientos de aplicación.  Pruebas de hipótesis con aplicaciones en distribuciones de: Medias, Proporciones.  Teoría de las muestras pequeñas. Distribución “t” de

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Answer:

3/5

Step-by-step explanation:

You divide 1 and 3. Keep the denominator since it is the same number

Hope this was helpful, Have a Great Day!!

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A team has P points.
Llana [10]
Sub in the values and solve for W
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4 years ago
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Jean works for the government and was conducting a survey to determine the income levels of a number of different neighborhoods
Brrunno [24]

Answer:

0.0418 = 4.18% probability that the average income level in the neighborhoods was less than $38,000.

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 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 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.

Jean knows that the mean income level in the country is $40,000, with a standard deviation of $2,000.

This means that \mu = 40000, \sigma = 2000

Jean selected three neighborhoods and determined the average income level.

This means that n = 3, s = \frac{2000}{\sqrt{3}} = 1154.7

What is the probability that the average income level in the neighborhoods was less than $38,000

This is the pvalue of Z when X = 38000. So

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

By the Central Limit Theorem

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

Z = \frac{38000 - 40000}{1154.7}

Z = -1.73

Z = -1.73 has a pvalue of 0.0418

0.0418 = 4.18% probability that the average income level in the neighborhoods was less than $38,000.

7 0
3 years ago
There are 10 kids in class. 7 boys and 3 girls. If the teacher were to pick 4 students at random what is the probability of sele
inysia [295]
Well there are 7 boys and 10 students total so 7/10=0.7=70%. Therefore there is a 70% chance.
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