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Instant Dinner comes in packages with weights that are normally distributed, with a standard deviation of 0.5 oz. If 2.3% of the dinners weigh more than 13.2 oz, what is the mean weight?

Respuesta :

Answer:

12.2

Step-by-step explanation:

2.3%= 0.023

So that is the probability that it will be over 13.2 oz, or .977 is the prob that it will be below 13.2

z-score = (real score - mean)/standard deviation

I found .977 at a z-score of 2.00

so 2.00 = (13.2-m)/0.5

=> m=12.2