I need help on this question

Si X es una variable aleatoria de tipo continuo… ¿Es posible definir el índice de asimetría de Fisher? ¿Por qué?

Nana76 [90]

Answer:

Primero, para un caso discreto, el índice de asimetría de Fisher se calcula como:

$\frac{\sum(xi - M)*n_i}{N*S_x}$

Donde:

xi son los i-esimos distintos valores de nuestro conjunto de datos.

M es la media

N es el numero de datos

ni es la frequencia absoluta de cada xi

Sx es la desviacion tipica

Podemos ver que en el caso discreto, escribimos el coeficiente con una sumatoria, lo cual realmente no podemos hacer para una variable continua.

Lo que si podemos hacer para una variable continua, es una integral, donde como esta vez tenemos una variable continua, podemos definir una función:

f(X) = frequencia absoluta para X.

Y escribir el coeficiente como:

$\frac{\int\limits^a_b {(X - M)*f(x)} \, dx }{N*S_x}$

Entonces si, es posible definir el índice.

8 0

3 years ago

You currently pay $10,400 a year for grounds maintenance. it is predicted that competition in the marketplace will force a 3% re

motikmotik

Answer:c= the price will be of $9, 492

Step-by-step explanation:

Hope this helps

4 0

3 years ago

Rosa works in an office building with many floors.From her office she goes down 5 floors for a meeting .Then she goes up 8 floor

n200080 [17]

Answer:

11th floor

Step-by-step explanation:

So to figure out what floor Rosa's office is in, we need to declare some variables.

x = Rosa's office

down 5 floors = -5

up 8 floors = 8

down 13 floors = - 13

One major clue to find Rosa's office is that she went down 13 floors to get to the 1st floor. So let's declare that Rosa is on the first floor.

x - 5 + 8 - 13 = 1

x - 10 = 1

x = 1 + 10

x = 11

So Rosa's office is at the 11th floor of the building.

8 0

3 years ago

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$