Question

the lengths of a lawn mower part are approximately normally distributed with a given mean \\( \mu = 4 \\) in and standard deviation \\( \sigma = 0.2 \\) in. what percentage of the parts will have lengths between 3.8 in. and 4.2 in?\
\\( \bigcirc \\) 34%\
\\( \bigcirc \\) 68%\
\\( \bigcirc \\) 95%\
\\( \bigcirc \\) 99.7%

Explanation:

Step1: Identify the number of standard deviations

First, we calculate how many standard deviations away 3.8 and 4.2 are from the mean (\(\mu = 4\)) with standard deviation (\(\sigma = 0.2\)).

For \(3.8\): \( \frac{3.8 - 4}{0.2} = \frac{-0.2}{0.2} = -1\) (so 3.8 is 1 standard deviation below the mean).

For \(4.2\): \( \frac{4.2 - 4}{0.2} = \frac{0.2}{0.2} = 1\) (so 4.2 is 1 standard deviation above the mean).

Step2: Apply the empirical rule for normal distribution

The empirical rule (or 68 - 95 - 99.7 rule) for a normal distribution states that approximately 68% of the data lies within 1 standard deviation (\(\mu \pm \sigma\)) of the mean, 95% within 2 standard deviations (\(\mu \pm 2\sigma\)), and 99.7% within 3 standard deviations (\(\mu \pm 3\sigma\)) of the mean.

Since we are looking at the range within 1 standard deviation of the mean (\(3.8 = 4 - 0.2\) and \(4.2 = 4 + 0.2\), which is \(\mu \pm \sigma\)), the percentage of data in this range is approximately 68%.

Answer:

68% (the option corresponding to 68%)