A sample of labor cost per hour to assemble a certain product has a mean of $2.60 and a standard deviation of 0.15. Use Chebyshev's theorem to find the values in which at least 88.89% of the data will lie.

Chebyshev's theorem states that for any dataset, the proportion of values that lie within $ k $ standard deviations of the mean is at least $ 1 - \frac{1}{k^2} $. You are given that at least 88.89% of the data is within a certain number of standard deviations from the mean. We can use this to find the value of $ k $.

### Step 1: Set up the equation using Chebyshev's theorem
Since 88.89% of the data lies within $ k $ standard deviations of the mean, we have:
\[
1 - \frac{1}{k^2} = 0.8889
\]

### Step 2: Solve for $ k $
\[
\frac{1}{k^2} = 1 - 0.8889 = 0.1111
\]
\[
k^2 = \frac{1}{0.1111} \approx 9
\]
\[
k = \sqrt{9} = 3
\]

So, at least 88.89% of the data lies within 3 standard deviations of the mean.

### Step 3: Calculate the range of values
The mean is $2.60, and the standard deviation is $0.15. Using the value $ k = 3 $, we calculate the range as follows:

\[
\text{Lower bound} = \mu - 3\sigma = 2.60 - 3(0.15) = 2.60 - 0.45 = 2.15
\]
\[
\text{Upper bound} = \mu + 3\sigma = 2.60 + 3(0.15) = 2.60 + 0.45 = 3.05
\]

### Conclusion:
At least 88.89% of the data will lie between **$2.15** and **$3.05**.

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Would you like further details or have any questions about this process?

Here are 5 related questions to explore:
1. How does Chebyshev's theorem compare with the empirical rule?
2. Can Chebyshev's theorem be used for any distribution?
3. How does increasing the percentage of data affect the range in Chebyshev’s theorem?
4. What are the limitations of using Chebyshev's theorem?
5. How would the calculation change if the standard deviation were 0.20 instead of 0.15?

**Tip**: Chebyshev's theorem is useful when you don't know the shape of the distribution, as it works for any distribution type!

Discover how to apply Chebyshev's theorem to determine the range where at least 88.89% of labor cost data lies. With a mean of $2.60 and a standard deviation of $0.15, explore the calculations that show the data falls between $2.15 and $3.05. Perfect for high school students studying statistics.

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