Given that the weights of farmer carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1 ounces.
The probability of a normally distributed data between two values (a, b) is given by:
[tex]P(a\ \textless \ X\ \textless \ b)=P\left(z\ \textless \ \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma/\sqrt{n}} \right) \\ \\ =P(7.5\ \textless \ X\ \textless \ 8.5)=P\left(z\ \textless \ \frac{8.5-8.0}{1.1/\sqrt{6}} \right)-P\left(z\ \textless \ \frac{7.5-8.0}{1.1/\sqrt{6}} \right) \\ \\ =P(z\ \textless \ 1.113)-P(-1.113)=0.8672-0.1328=0.7344[/tex]
