Answer:
Step-by-step explanation:
From the given information:
the mean [tex](\mu) = 115 \times 20[/tex]
= 2300
Standard deviation = [tex]20 \times \sqrt{115}[/tex]
Standard deviation (SD) = 214.4761
TO find:
a) [tex]P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})[/tex]
[tex]P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})[/tex]
[tex]P(x > 3500)= P(Z > \dfrac{1200}{214.4761})[/tex]
[tex]P(x > 3500)= P(Z >5.595)[/tex]
From the Z-table, since 5.595 is > 3.999
[tex]P(x > 3500)=1-0.9999[/tex]
P(x > 3500) = 0.0001
b)
Here, the replacement time for the mean [tex](\mu) = \dfrac{0+0.5}{2}[/tex]
= 0.25
Replacement time for the Standard deviation [tex]\sigma = \dfrac{0.5-0}{\sqrt{12}}[/tex]
[tex]\sigma = 0.1443[/tex]
For 115 component, the mean time = (115 × 20)+(114×0.25)
= 2300 + 28.5
= 2328.5
Standard deviation = [tex]\sqrt{(115\times 20^2) +(114\times (0.1443)^2)}[/tex]
= [tex]\sqrt{(115\times 400) +(114\times 0.02082249}[/tex]
= [tex]\sqrt{(46000) +2.37376386}[/tex]
= [tex]\sqrt{(46000) +(2.37376386)}[/tex]
= [tex]\sqrt{46002.374}[/tex]
= 214.482
Now; the required probability:
[tex]P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})[/tex]
[tex]P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})[/tex]
[tex]P(x > 4125) = P(Z >8.376)[/tex]
[tex]P(x > 4125) =1- P(Z <8.376)[/tex]
From the Z-table, since 8.376 is > 3.999
P(x > 4125) = 1 - 0.9999
P(x > 4125) = 0.0001
