Hi, I am trying to solve a problem using Chebyshev's Theorem. The problem says that:
A large sample of Northern Pike caught at Taltson Lake (Canada) showed that the average length was x (mean)=32.5 inches with sample standard deviation s=8.6 inches.
a.) Compute the coefficient of variation for this data. (The coefficient of variation is 26.26)
b.) Use Chebyshev's Theorem to find an interval centered on the mean in which we can expect at least 75% of the data to fall.
I don't understand Part B. Can someone please explain to me how to find the interval using Chebyshev's Theorem?
From a sample with equals=28, the mean number of televisions per household is 2 with a standard deviation of 11 television. Using Chebychev's Theorem, determine at least how many of the households have between 0 and 4 televisions.
To use Chebyshev's Theorem to find an interval centered on the mean in which we can expect at least 75% of the data to fall, follow these steps:
Step 1: Define the coefficient of variation (CV). The coefficient of variation is given as 26.26, which is calculated by dividing the sample standard deviation (s) by the mean (x) and multiplying by 100.
CV = (s / x) * 100 = (8.6 / 32.5) * 100 = 26.46 (approximately)
Step 2: Use the coefficient of variation to find the lower bound on the standard deviation. The lower bound (LB) is calculated by dividing 100 by the coefficient of variation.
LB = 100 / CV = 100 / 26.46 ≈ 3.78
Note: The lower bound represents the minimum number of standard deviations away from the mean within which at least 75% of the data falls.
Step 3: Apply Chebyshev's Theorem. According to Chebyshev's Theorem, at least (1 - 1 / k^2) * 100% of the data will fall within k standard deviations of the mean, where k is a positive integer greater than 1. In this case, we want at least 75% of the data to fall within the interval, so we set (1 - 1 / k^2) equal to 0.75.
1 - 1 / k^2 = 0.75
Solving for k:
1 / k^2 = 0.25
k^2 = 1 / 0.25
k^2 = 4
k = sqrt(4)
k = 2
Step 4: Calculate the interval. Multiply the lower bound by the standard deviation to get the interval.
Interval = LB * s = 3.78 * 8.6 = 32.508 (approximately)
Therefore, we can expect at least 75% of the data to fall within an interval of 32.508 inches centered on the mean.
Sure! I can help you with that.
According to Chebyshev's Theorem, for any distribution (regardless of its shape), the proportion of data falling within a certain number of standard deviations from the mean will be at least (1 - 1/z^2), where z is the number of standard deviations.
In your case, you want to find an interval centered on the mean within which at least 75% of the data will fall.
To apply Chebyshev's Theorem, we need to determine the number of standard deviations, denoted by z. The formula is:
z = (x - μ) / σ,
where x is the number of standard deviations away from the mean, μ is the mean, and σ is the standard deviation.
In this case, you want at least 75% of the data to fall within the interval. So the proportion outside the interval will be (1 - 0.75) = 0.25.
Now, substitute the given values into the formula:
z = (0.25 / 2) = 0.125.
This means that at least 75% of the data will fall within 0.125 standard deviations from the mean.
To get the interval, multiply the standard deviation (s) by the number of standard deviations (z):
Interval = z * s = 0.125 * 8.6 inches.
Therefore, the interval centered on the mean in which we can expect at least 75% of the data to fall is approximately ±1.075 inches from the mean value of 32.5 inches.
Hope this helps! Let me know if you have any further questions.