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

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

the lengths of a lawn mower part are approximately normally distributed with a given mean μ = 4 in. and standard deviation σ = 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

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

The empirical rule for a normal distribution states that approximately 68% of the data lies within 1 standard - deviation of the mean. Since $z_1=-1$ and $z_2 = 1$, the percentage of parts with lengths between 3.8 in. and 4.2 in. is 68%.

Answer:

68%