the lengths of a lawn mower part are approximately normally distributed with a given mean $mu = 4$ in. and…

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.?\n34%\n68%\n95%\n99.7%

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.?\n34%\n68%\n95%\n99.7%

Answer

Answer:

B. 68%

Explanation:

Step1: Calculate z - scores

For (x_1 = 3.8), (z_1=\frac{3.8 - 4}{0.2}=\frac{- 0.2}{0.2}=-1). For (x_2 = 4.2), (z_2=\frac{4.2 - 4}{0.2}=\frac{0.2}{0.2}=1).

Step2: Use empirical rule

In a normal - distribution, approximately 68% of the data lies within 1 standard deviation ((\mu\pm\sigma)) of the mean. Here, the interval ((3.8,4.2)) is (\mu\pm\sigma) since (\mu = 4) and (\sigma=0.2) ((4 - 0.2=3.8) and (4 + 0.2 = 4.2)). So the percentage of parts with lengths between 3.8 in. and 4.2 in. is 68%.