Solution: We are given:
[tex]\mu=0.33, \sigma =0.7[/tex]
Let [tex]x[/tex] be the weight (oz) of laptop
We have to find [tex]P(x>32)[/tex]
To find the this probability, we need to find the z score value.
The z score is given below:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]=\frac{32-33}{0.7}[/tex]
[tex]=-1.43[/tex]
Now, we have to find [tex]P(z>-1.43)[/tex]
Using the standard normal table, we have:
[tex]P(z>-1.43)=0.9236[/tex]
0.9236 or 92.36% of laptops are overweight