Question

The average life of Canadian women is 73.75 years, and the standard deviation of the life expectancy of Canadian women is 6.5 years. Based on Chebyshev's theorem, determine the upper and lower bounds on the average life expectancy of Canadian women such that at least 90 percent of the population is included. a. 53.20; 94.30b. 66.38, 81.13 c. 67.25, 80.25 d. 12.09, 135.41

Answer 1

Using Chebyshev's theorem to find the range of life expectancy for Canadian women that includes at least 90% of the population, the lower and upper bounds are approximately 53.21 years and 94.29 years, respectively.

Step-by-step explanation:

The question pertains to calculating bounds on life expectancy using Chebyshev's theorem, which is a statistical rule that applies to different types of distributions, regardless of their shape. To calculate the bounds that include at least 90% of the data for the life expectancy of Canadian women, where the mean is 73.75 years, and the standard deviation is 6.5 years, we need to use the formula k = 1/√(1-(1/p)), where p is the proportion of the population. In this case, since we want to include at least 90% of the population, p=0.9.

First, we solve for k:

k = 1/√(1-(1/0.9))
k ≈ 3.16

Then, we multiply k by the standard deviation and subtract it from and add it to the mean to find the bounds:

Lower Bound = Mean - k * Standard Deviation
Lower Bound = 73.75 - (3.16 * 6.5) ≈ 53.21

Upper Bound = Mean + k * Standard Deviation
Upper Bound = 73.75 + (3.16 * 6.5) ≈ 94.29

Therefore, the bounds are approximately 53.21 years and 94.29 years, which corresponds to option a. 53.20; 94.30.

Answer 2

a. 53.20; 94.30

Step-by-step explanation:

Data given

`\mu =73.75` reprsent the population mean

`\sigma=6.5` 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-(1)/(k^2)`

We want the limits that have at least 90% of the population is include. And using the theorem we have this:

`0.9 =1-(1)/(k^2)`

And solving for k we have this:

`(1)/(k^2)=0.1`

`k^2 =(1)/(0.1)=10`

`k=\pm 3.162`

So then we need the limits between two deviations from the mean in order to have at least 90% of the data will reside.

Lower bound:

`\mu -3.162\sigma=73.75-3.162(6.5)=53.195 \apporx 53.20`

Upper bound:

`\mu +3.192\sigma=73.75+3.162(6.5)=94.304 \approx 94.30`

So the final answer would be between (53.20;94.30)