Question

from a sample with n = 32, the mean duration of a geyser’s eruptions is 3.68 minutes and the standard deviation is 0.54 minutes. using chebychev’s theorem, determine at least how many of the eruptions lasted between 2.6 and 4.76 minutes.

at least \\(\square\\) of the eruptions lasted between 2.6 and 4.76 minutes.
(simplify your answer.)

Explanation:

Step1: Identify the mean and k for Chebyshev's

The mean ($\mu$) is 3.68, and we need to find how many standard deviations ($k$) 2.6 and 4.76 are from the mean. First, calculate the distance from the mean: $3.68 - 2.6 = 1.08$ and $4.76 - 3.68 = 1.08$. So the interval is $\mu \pm k\sigma$, where $k\sigma = 1.08$. Given $\sigma = 0.54$, solve for $k$: $k = \frac{1.08}{0.54} = 2$.

Step2: Apply Chebyshev's Theorem

Chebyshev's Theorem states that at least $1 - \frac{1}{k^2}$ of the data lies within $k$ standard deviations of the mean. For $k = 2$, this is $1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$.

Step3: Calculate the number of eruptions

The sample size $n = 32$. Multiply the proportion by $n$: $\frac{3}{4} \times 32 = 24$.

Answer:

24