Answer: 0.9871
Step-by-step explanation:
Given : A manufacturer knows that their items have a lengths that are approximately normally distributed with
[tex]\mu=5.8 \text{ inches}[/tex]
[tex]\sigma=0.5\text{ inches}[/tex]
Sample size : [tex]n=31[/tex]
Let x be the length of randomly selected item.
z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]z=\dfrac{5.6-5.8}{\dfrac{0.5}{\sqrt{31}}}\approx-2.23[/tex]
The probability that their mean length is greater than 5.6 inches by using the standard normal distribution table
= [tex]P(x>5.6)=P(z>-2.23)=1-P(z<-2.23)[/tex]
[tex]=1-0.0128737=0.9871263\approx0.9871[/tex]
Hence, the probability that their mean length is greater than 5.6 inches is 0.9871.
