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 from the mean

First, we calculate how many standard deviations 3.8 and 4.2 are from the mean (\(\mu = 4\), \(\sigma = 0.2\)).
For \(3.8\): \(4 - 3.8 = 0.2\), so \(3.8=\mu - \sigma\) (since \(0.2=\sigma\)).
For \(4.2\): \(4.2 - 4 = 0.2\), so \(4.2=\mu + \sigma\).

Step2: Apply the empirical rule for normal distribution

The empirical rule (68 - 95 - 99.7 rule) states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(\mu \pm \sigma\) (1 standard deviation from the mean).
• Approximately \(95\%\) lies within \(\mu \pm 2\sigma\).
• Approximately \(99.7\%\) lies within \(\mu \pm 3\sigma\).

Since our interval is \(\mu - \sigma\) to \(\mu + \sigma\), the percentage of data within this range is \(68\%\).

Answer:

68% (corresponding to the option with "68%")