Question

A real estate company is interested in the ages of home buyers. They examined the ages of thousands of home buyers and found that the mean age was 49 years old. Suppose that this measure is valid for the population of all home buyers. Complete the following statements about the distribution of all ages of home buyers.(a) According to Chebyshev's theorem, at least 56 75 84 89 of the home buyers' ages lie within 2.5 standard deviations of the mean, years.(b) Suppose that the distribution is bell-shaped. If approximately 68% of the home buyers' ages lie between 39 years and 59 years, then the approximate value of the standard deviation for the distribution, according to the empirical rule, is .

Answer 1

Answer:

a) $1- \frac{1}{2.5^2} = 0.84$

And that represent 84% of the data within 2.5 deviations from the mean

b) For this case we can assume that the limits between 39 and 59 are given by:

$39 =\mu -\sigma= 49-\sigma$

$59 =\mu +\sigma= 49+\sigma$

Because within one deviation from the mena we have at least 68% of the data.

And we can solve for the deviation and we got:

$\sigma = 49-39 = 10$

$\sigma= 59-49 = 10$

Step-by-step explanation:

Part a

Data given  

$\mu=49$ reprsent the population mean

$\sigma$ represent the population standard deviation

The Chebyshev's Theorem states that for any dataset

• We have at least 75% of all the data within two deviations from the mean.

• We have at least 88.9% of all the data within three deviations from the mean.

• We have at least 93.8% of all the data within four deviations from the mean.

Or in general words "For any set of data (either population or sample) and for any constant k greater than 1, the proportion of the data that must lie within k standard deviations on either side of the mean is at least: $1-\frac{1}{k^2}"$

And if we use the value of k=2.5 we got:

$1- \frac{1}{2.5^2} = 0.84$

And that represent 84% of the data within 2.5 deviations from the mean

Part b

For this case we can assume that the limits between 39 and 59 are given by:

$39 =\mu -\sigma= 49-\sigma$

$59 =\mu +\sigma= 49+\sigma$

Because within one deviation from the mena we have at least 68% of the data.

And we can solve for the deviation and we got:

$\sigma = 49-39 = 10$

$\sigma= 59-49 = 10$