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A company sells cookies in 250-gram packs. When 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. Assuming the data is normally distributed, we can conclude that % of the packs weighed less than 250 grams

Sagot :

Apologiabiologyidn
so the deviation was 2.5 grams
that means that the average/mean could have been
255-2.5 or 255+2.5 =   252.5 or 257.5
so if all of them weighed the same, none of them weighed less than 250 so the answer is 0%

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.

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