Question

What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United States gave the following information about a prehistoric extended family group of 88 members on what is now a Native American reservation. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint. (Round your answers to one decimal place.) Age range (years) 1-10 11-20 21-30 31 and over Number of individuals 40 15 23 10

Answer 1

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:

Answer 2

Answer:

$\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8$

$s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3$

$Sd(X) = \sqrt{118.273}=10.9$

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:

$\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{1394}{88}=15.8$

The variance owuld be given by this formula"

$s^2 = \frac{n(\sum x^2 f) -(\sum xf)^2}{n(n-1}$

And if we replace we got:

$s^2 = \frac{88(32372) -(1394)^2}{88(88-1}=118.3$

The standard deviation would be just the square root of the variance:

$Sd(X) = \sqrt{118.273}=10.9$