Answer:
9.248 < [tex]\mu[/tex] < 9.253 mm
Explanation:
Given data:
standard deviation = 0.021 mm
sample mean = 9.251 mm
total sample = 409 m
confidence interval level = 95%
mean diameter of entire batch [tex]\mu = \bar X \pm z \frac{\sigma}{\sqrt{n}}[/tex]
where
[tex] \bar X [/tex] - sample mean = 9.251 mm
z = critical value
plugginf all value in the above relation to get the mean diameter
[tex]\mu = 9.251 \pm 1.96 \times \frac{0.021}{\sqrt{409}}[/tex]
9.248 < [tex]\mu[/tex] < 9.253 mm
