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The weights of bags of ready-to-eat salad are normally distributed with a mean of 290 grams and a standard deviation of 10 grams.


What percent of the bags weigh less than 280 grams?


13.5%


16%


34%


36.5%

Verified Answer :

Answer: Second option is correct.

Step-by-step explanation:

Since we have given that

Mean = 290 grams

Standard deviation = 10 grams

Since the weight of bags of ready-to-eat salad are normally distributed.

We need to find the percent of the bags weigh less than 280 grams.

As we know that

\(z=\frac{X-\mu}{\sigma}\\\\z=\frac{280-290}{10}\\\\z=-10\)

\(P(X<280)=P(z<-1)= 0.1587\) (Using the normally distribution table)

If we change it into percentage.

\(0.1587\times 100\\\\=15.87\%\\\\\approx 16\%\)

Hence, Second option is correct.