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%

Explanation:

Step1: Calculate z - scores

For $x_1 = 3.8$, $z_1=\frac{x_1-\mu}{\sigma}=\frac{3.8 - 4}{0.2}=\frac{- 0.2}{0.2}=-1$.
For $x_2 = 4.2$, $z_2=\frac{x_2-\mu}{\sigma}=\frac{4.2 - 4}{0.2}=\frac{0.2}{0.2}=1$.

Step2: Use the empirical rule

The empirical rule for a normal distribution states that approximately 68% of the data lies within 1 standard - deviation of the mean. That is, $P(-1

Answer:

B. 68%