A certain geyser was determined to have eruptions with a Mean duration of 3.31 min and a standard deviation of 1.04 min. A sample of n=43 eruptions were recorded. Using Chebychev's Theorem, how many of the eruptions lasted between 1.23 min and 5.39 min.
there both 3 sd's away but i have no clue how to do this
To use Chebyshev's Theorem to find the proportion of eruptions that lasted between 1.23 min and 5.39 min, we need to calculate the range of the data within a certain number of standard deviations from the mean.
Chebyshev's Theorem states that for any number k (greater than 1), at least 1 - 1/k^2 of the data will fall within k standard deviations of the mean.
In this case, we need to find the proportion within 3 standard deviations from the mean.
Step 1: Calculate the range within 3 standard deviations of the mean.
To find the range, we multiply the standard deviation by the number of standard deviations, which gives us:
3 x 1.04 min = 3.12 min
Step 2: Calculate the range required for the data to fall within 1.23 min and 5.39 min.
To find the required range, we subtract the lower value from the higher value, which gives us:
5.39 min - 1.23 min = 4.16 min
Step 3: Calculate the proportion using Chebyshev's Theorem.
To find the proportion, we divide the required range by the range within 3 standard deviations:
Proportion = Required range / Range within 3 standard deviations
Proportion = 4.16 min / 3.12 min
Proportion ≈ 1.33
Since Chebyshev's Theorem guarantees that at least 1 - 1/k^2 of the data falls within k standard deviations from the mean, the proportion of eruptions that lasted between 1.23 min and 5.39 min will be at least 1 - 1/1.33^2.
Step 4: Calculate the proportion for the given data.
Proportion = 1 - 1/1.33^2
Proportion ≈ 1 - 1/1.7729
Proportion ≈ 1 - 0.5645
Proportion ≈ 0.4355
Therefore, at least 43.55% or 43.55% of the eruptions will last between 1.23 min and 5.39 min, according to Chebyshev's Theorem.
To use Chebyshev's Theorem, you need to calculate the proportion of data that lies within a certain number of standard deviations from the mean. Chebyshev's Theorem states that no matter the shape of the distribution, at least (1 - 1/z^2) of the data lies within z standard deviations from the mean, where z is any positive number greater than 1.
In your case, you want to find the proportion of eruptions that lasts between 1.23 min and 5.39 min, which is 3 standard deviations away from the mean on both sides. So, you can calculate the proportion using the following formula:
Proportion = 1 - (1/z^2)
To find z, you will need to standardize the values of 1.23 min and 5.39 min by subtracting the mean and dividing by the standard deviation.
Z1 = (1.23 - mean) / standard deviation
Z2 = (5.39 - mean) / standard deviation
Once you have the values of Z1 and Z2, you can find the proportion using Chebyshev's Theorem. Remember that the proportion represents the minimum percentage of data that meets the criteria.
So, the proportion of eruptions that lasted between 1.23 min and 5.39 min is:
Proportion = 1 - (1/z^2)
Proportion = 1 - (1/Z1^2) - (1/Z2^2)
However, Chebyshev's Theorem gives a lower bound estimate of the proportion. It tells you that at least this proportion meets the criteria, but it does not provide the exact percentage.