Respuesta :
Answer:
 We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .
Step-by-step explanation:
Given -
The sample size is large then we can use central limit theorem
n = 50 , Â
Standard deviation[tex](\sigma)[/tex] = 7.1
Mean [tex]\overline{(y)}[/tex] = 110
[tex]\alpha =[/tex] 1 - confidence interval = 1 - .98 = .02
[tex]z_{\frac{\alpha}{2}}[/tex] = 2.33
98% confidence interval for the mean caffeine content for cups dispensed by the machine = [tex]\overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}[/tex]
           = [tex]110\pm z_{.01}\frac{7.1}\sqrt{50}[/tex]
           = [tex]110\pm 2.33\frac{7.1}\sqrt{50}[/tex]
    First we take  + sign
  [tex]110 + 2.33\frac{7.1}\sqrt{50}[/tex] = 112.34
now  we take  - sign
[tex]110 - 2.33\frac{7.1}\sqrt{50}[/tex] = 107.66
 We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .
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