Respuesta :
Answer:
68.26% of the parts will have lengths between 3.8 in. and 4.2 in.
Step-by-step explanation:
The lengths of a lawn mower part are approximately normally distributed
[tex]\mu = 4[/tex]
Standard deviation =[tex]\sigma =0.2[/tex]
We are supposed to find What percentage of the parts will have lengths between 3.8 in. and 4.2 in . i.e. P(3.8<x<4.2)
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
At x = 3.8
[tex]Z=\frac{3.8-4}{0.2}[/tex]
Z=-1
Refer the z table for p value
p value =0.1587
At x =4.2
[tex]Z=\frac{4.2-4}{0.2}[/tex]
Z=1
Refer the z table for p value
p value =0.8413
P(3.8<x<4.2)=P(x<4.2)-P(x<3.8)=0.8413-0.1587=0.6826
So, 68.26% of the parts will have lengths between 3.8 in. and 4.2 in.
