Question

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%

Answer 1

Answer: Second option is correct.

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=(X-\mu)/(\sigma)\\\\z=(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* 100\\\\=15.87\%\\\\\approx 16\%`

Hence, Second option is correct.