Respuesta :
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8[/tex]
[tex] s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3[/tex]
[tex]Sd(X) = \sqrt{118.273}=10.9[/tex]
Step-by-step explanation:
Previous concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
For this case we can calculate the properties required with the following table:
Interval Mid point (x) f x*f x^2 *f
_________________________________________
1-10 5.5 40 220 1210
11-20 15.5 15 232.5 3603.75
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
________________________________________
Total 88 1394 32372
We assume that the mid point for the class >31 is 35.5 using the problem information.
For this case the expected value would be given by:
[tex] \bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8[/tex]
The variance owuld be given by this formula"
[tex] s^2 = \frac{n(\sum x^2 f) -(\sum xf)^2}{n(n-1}[/tex]
And if we replace we got:
[tex] s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3[/tex]
The standard deviation would be just the square root of the variance:
[tex]Sd(X) = \sqrt{118.273}=10.9[/tex]
Answer:
revious concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
For this case we can calculate the properties required with the following table:
Interval Mid point (x) f x*f x^2 *f
_________________________________________
1-10 5.5 40 220 1210
11-20 15.5 15 232.5 3603.75
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
________________________________________
Total 88 1394 32372
We assume that the mid point for the class >31 is 35.5 using the problem information.
For this case the expected value would be given by:
The variance owuld be given by this formula"
And if we replace we got:
The standard deviation would be just the square root of the variance:
revious concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
For this case we can calculate the properties required with the following table:
Interval Mid point (x) f x*f x^2 *f
_________________________________________
1-10 5.5 40 220 1210
11-20 15.5 15 232.5 3603.75
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
________________________________________
Total 88 1394 32372
We assume that the mid point for the class >31 is 35.5 using the problem information.
For this case the expected value would be given by:
The variance owuld be given by this formula"
And if we replace we got:
The standard deviation would be just the square root of the variance:
Step-by-step explanation:
