Answer:
The probability that the sample mean would differ from the population mean by greater than 2.2 kilograms is 0.0104 .
Step-by-step explanation:
The mean weight of an adult is 62 kilograms with a variance of 144
i.e. [tex]\mu = 62 \\\sigma^2 = 144[/tex]
We are supposed to find probability that the sample mean would differ from the population mean by greater than 2.2 kilograms
i.e. [tex]P(\bar{x}<62-2.2) or P(\bar{x}>62+2.2)=1-P(59.8<\bar{x}<64.2)[/tex]
Using formula : [tex]\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]P(\bar{x}<62-2.2) or P(\bar{x}>62+2.2)=1-P(59.8<\bar{x}<64.2)[/tex]
[tex]P(\bar{x}<62-2.2) or P(\bar{x}>62+2.2)=1-P(\frac{59.8-62}{\frac{12}{\sqrt{195}}}<\frac{64.2-62}{\frac{12}{\sqrt{195}}})[/tex]
[tex]P(\bar{x}<62-2.2) or P(\bar{x}>62+2.2)=1-P(-2.56<z<2.56)[/tex]
[tex]P(\bar{x}<62-2.2) or P(\bar{x}>62+2.2)=1-{P(z<2.56)-P(z<-2.56)}[/tex]
Refer the z table
[tex]P(\bar{x}<62-2.2) or P(\bar{x}>62+2.2)=1-{0.9948-0.0052}[/tex]
[tex]P(\bar{x}<62-2.2) or P(\bar{x}>62+2.2)=0.0104[/tex]
Hence The probability that the sample mean would differ from the population mean by greater than 2.2 kilograms is 0.0104 .