Answer:
2.28%
Step-by-step explanation:
We have been given that a particular batch of 1,000 packs was weighed, the mean weight per pack was 255 grams and the standard deviation was 2.5 grams.
We will use z-score formula to solve our given problem.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Raw score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex]
Upon substituting our given values in z-score formula we will get,
[tex]z=\frac{250-255}{2.5}[/tex]
[tex]z=\frac{-5}{2.5}[/tex]
[tex]z=-2[/tex]
Now we will use normal distribution table to find the area corresponding to z-score of -2.
Using normal distribution table we will get,
[tex]P(z<-2)=0.02275[/tex]
To convert our answer to percentage we will multiply 0.02275 by 100.
[tex]\text{Percentage of packs weighed less than 250 grams}=0.02275\times 100[/tex]
[tex]\text{Percentage of packs weighed less than 250 grams}=2.275\%\approx 2.28\%[/tex]
Therefore, 2.28% of the packs weighed less than 250 grams.
