Question

from a sample with n = 20, the mean number of pets per household is 3 with a standard deviation of 1 pet. using chebychevs theorem, determine at least how many of the households have 1 to 5 pets. at least of the households have 1 to 5 pets. (simplify your answer)

Explanation:

Step1: Recall Chebyshev's Theorem

Chebyshev's Theorem states that for any number \(k> 1\), the proportion of the data that lies within \(k\) standard - deviations of the mean is at least \(1-\frac{1}{k^{2}}\). First, we need to find \(k\) such that the interval \([\mu - k\sigma,\mu + k\sigma]\) corresponds to the interval \([1,5]\) given \(\mu = 3\) and \(\sigma = 1\).

Step2: Calculate \(k\) values

We set up the equations \(\mu - k\sigma=1\) and \(\mu + k\sigma = 5\). Substituting \(\mu = 3\) and \(\sigma = 1\) into \(\mu - k\sigma=1\), we get \(3 - k\times1=1\), which gives \(k = 2\). Similarly, substituting into \(\mu + k\sigma=5\) gives \(3 + k\times1=5\), which also gives \(k = 2\).

Step3: Apply Chebyshev's Theorem

Using the formula \(1-\frac{1}{k^{2}}\), with \(k = 2\), we have \(1-\frac{1}{2^{2}}=1-\frac{1}{4}=\frac{3}{4}=0.75\). If we have a sample of \(n\) households, the number of households with 1 to 5 pets is at least \(0.75n\).

Answer:

If the sample size is \(n\), at least \(0.75n\) households have 1 to 5 pets. If we assume \(n = 20\) (since it was mentioned \(n = 20\) in a non - clear part of the problem statement), then at least \(0.75\times20 = 15\) households have 1 to 5 pets.