Question

from a sample with ( n = 36 ), the mean duration of a geysers eruptions is 3.61 minutes and the standard deviation is 0.94 minutes. using chebychevs theorem, determine at least how many of the eruptions lasted between 1.73 and 5.49 minutes.
at least ( square ) of the eruptions lasted between 1.73 and 5.49 minutes.
(simplify your answer)

Explanation:

Step1: Calculate the number of standard deviations \(k\)

The formula for \(k\) is \(k=\frac{\text{Value}-\mu}{\sigma}\).
For the lower bound: \(k_1=\frac{3.61 - 1.73}{0.94}=\frac{1.88}{0.94} = 2\)
For the upper bound: \(k_2=\frac{5.49 - 3.61}{0.94}=\frac{1.88}{0.94}=2\)

Step2: Apply Chebyshev's Theorem

Chebyshev's Theorem states that \(P(|X-\mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}\)
Substitute \(k = 2\) into the formula: \(1-\frac{1}{2^{2}}=1-\frac{1}{4}=\frac{3}{4}\)

Step3: Calculate the number of data points

The number of data points \(n = 36\)
The number of eruptions is \(n\times(1-\frac{1}{k^{2}})\)
Substitute \(n = 36\) and \(1-\frac{1}{k^{2}}=\frac{3}{4}\)
\(36\times\frac{3}{4}=27\)

Answer:

\(27\)