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. ? 34% 68% 95% 99. 7%.

Answer 1

Empirical rule is defined for normal distributions. The percentage of the parts the lengths between 3.8 inches and 4.2 inches lie is: Option B: 68%

What is empirical rule?

According to the empirical rule, also known as 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95% and 99.7% of the values lies within one, two or three standard deviations of the mean of the distribution.

$P(\mu - \sigma < X < \mu + \sigma) = 68\%\\P(\mu - 2\sigma < X < \mu + 2\sigma) = 95\%\\P(\mu - 3\sigma < X < \mu + 3\sigma) = 99.7\%$

where we had  

$X \sim N(\mu, \sigma)$

where mean of distribution of X is $\mu$ and standard deviation from mean of distribution of X is $\sigma$

For the given case, as it is given that the lengths of a lawn mower part are approximately normally distributed, let its lengths be tracked by a random variable X

Then, $X \sim N(\mu =4, \sigma = 0.2)$

The needed percentage is

$P(3.8 < X < 4.2)$

We can rewrite it as:

$P(4-0.2 < X < 4 + 0.2)$

This is the percent of values lying under one standard deviation about the mean of X. Using the empirical rule, we get the percentage as:

$P(\mu - \sigma < X < \mu + \sigma) = 68\%$

Thus,

$P(4-0.2 < X < 4 + 0.2) = 68\%$

Thus, The percentage of the parts the lengths between 3.8 inches and 4.2 inches lie is: Option B: 68%

Learn more about empirical rule here:

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