Answer: 0.0078
Step-by-step explanation:
We assume that cans of Coke are filled so that the actual amounts are normally distributed with a mean of 12.00 oz and a standard deviation of 0.12 oz.
As per given , we have
[tex]\mu=12.00\ oz[/tex] and [tex]\sigma=0.12\ oz[/tex]
Sample size : n= 5
Let [tex]\overline{X}[/tex] be the sample mean amount of coke in can.
Then, the probability that a sample of 5 cans will have a mean amount of at least 12.13 oz will be :_
[tex]P(\overline{x}\geq12.13)=1-P(\overline{x}<12.13)\\\\=1-P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{12.13-12.00}{\dfrac{0.12}{\sqrt{5}}})[/tex]
[tex]=1-P(z<2.42)\ \ \ [\because \ z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}][/tex]
[tex]= 1-0.9922=0.0078[/tex] [By using z-table]
Hence, the required probability = 0.0078
