Answer: 0.2140
Step-by-step explanation:
Given : A company produces steel rods. The lengths of the steel rods are normally distributed with
[tex]\mu=92.6 \text{ cm}[/tex]
[tex]\sigma=2\text{ cm}[/tex]
Sample size : [tex]n=12[/tex]
Let x be the length of randomly selected item.
z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x=92 cm
[tex]z=\dfrac{92-92.6}{\dfrac{2}{\sqrt{12}}}\approx-1.04[/tex]
For x=92.4 cm
[tex]z=\dfrac{92.4-92.6}{\dfrac{2}{\sqrt{12}}}\approx-0.35[/tex]
The probability that the average length of a randomly selected bundle of steel rods is between 92-cm and 92.4-cm by using the standard normal distribution table
= [tex]P(92<x<92.4)=P(-1.04<z<-0.35)=P(z<-0.35)-P(z<-1.04)[/tex]
[tex]= 0.3631693-0.14917=0.2139993\approx0.2140[/tex]
Hence, the probability that the average length of a randomly selected bundle of steel rods is between 92-cm and 92.4-cm is 0.2140.
