The average life of Canadian women is 73.75 years and the standard deviation of the women's life expectancy in Canada is 6.5 years. Using the Chebyshev's theorem, determine the minimum percentage of women in Canada whose life expectancy is between 64 and 83.5 years. Round your answer to 2 decimal places and express as a percentage (%).
Some hints:
This theorem says
1. Within two standard deviations of the mean, you will find at least 75% of the data.
2. Within three standard deviations of the mean, you will find at least 89% of the data.
Here's how the formula shows this:
Formula is 1 - (1/k^2) ---> ^2 means squared.
If k = 2 (representing two standard deviations), we have this:
1 - (1/2^2) = 1 - (1/4) = 3/4 or .75 or 75%
If k = 3 (representing three standard deviations), we have this:
1 - (1/3^2) = 1 - (1/9) = 8/9 or approximately .89 or 89%
Looking at your problem, the difference between the mean and the data given is 1.5 standard deviations.
Therefore, we have this:
1 - (1/1.5^2) = 1 - (1/2.25) = 1 - 0.44 = 0.56 = 56%
I hope this helps.
asasdas
To determine the minimum percentage of women in Canada whose life expectancy is between 64 and 83.5 years using Chebyshev's theorem, we need to calculate the number of standard deviations away from the mean each of these values lies.
Let's start with the lower bound, which is 64 years. To find the number of standard deviations away from the mean, we subtract the mean (73.75) from the lower bound (64) and divide by the standard deviation (6.5):
z1 = (64 - 73.75) / 6.5 = -1.48
Next, we'll do the same for the upper bound, which is 83.5 years:
z2 = (83.5 - 73.75) / 6.5 = 1.5
Now that we have the number of standard deviations for both values, we can apply Chebyshev's theorem. According to Chebyshev's theorem, at least (1 - 1/z^2) of the data lies within z standard deviations from the mean.
For z1 = -1.48, the minimum percentage of data within this range is 1 - 1/(-1.48)^2.
For z2 = 1.5, the minimum percentage of data within this range is 1 - 1/(1.5)^2.
Calculating these values:
Percentage for z1 = 1 - 1/(-1.48)^2 = 0.7441 = 74.41%
Percentage for z2 = 1 - 1/(1.5)^2 = 0.7778 = 77.78%
Therefore, the minimum percentage of women in Canada whose life expectancy is between 64 and 83.5 years, according to Chebyshev's theorem, is 74.41% and 77.78% when rounded to 2 decimal places and expressed as a percentage.
To use Chebyshev's theorem to determine the minimum percentage of women in Canada whose life expectancy is between 64 and 83.5 years, we can use the inequality:
\[1 - \frac{1}{{k^2}} \leq \frac{{\text{{number of values in the range}}}}{{\text{{total number of values}}}} \leq 1 - \frac{1}{{k^2}}\]
where \(k\) is the number of standard deviations from the mean.
First, let's calculate the number of standard deviations for each endpoint of the range:
For the lower endpoint (64 years):
\(k = \frac{{\text{{value}} - \text{{mean}}}}{{\text{{standard deviation}}}} = \frac{{64 - 73.75}}{{6.5}} \approx -1.50\)
For the upper endpoint (83.5 years):
\(k = \frac{{\text{{value}} - \text{{mean}}}}{{\text{{standard deviation}}}} = \frac{{83.5 - 73.75}}{{6.5}} \approx 1.50\)
Now, we can determine the minimum percentage using \(k = 1.50\):
\[1 - \frac{1}{{(1.5)^2}} \leq \text{{percentage}} \leq 1 - \frac{1}{{(1.5)^2}}\]
Calculating this inequality:
\[1 - \frac{1}{{2.25}} \leq \text{{percentage}} \leq 1 - \frac{1}{{2.25}}\]
\[1 - \frac{4}{9} \leq \text{{percentage}} \leq 1 - \frac{4}{9}\]
\[\frac{5}{9} \leq \text{{percentage}} \leq \frac{5}{9}\]
Since the minimum value is equal to the maximum value, the minimum percentage of women in Canada whose life expectancy is between 64 and 83.5 years is \(\frac{5}{9}\) or approximately 0.56 rounded to two decimal places. In percentage form, this is approximately 56%.