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7 July, 21:51

The weight distribution of parcels sent in a certain manner is normal with mean value 15 lb and standard deviation 3.3 lb. The parcel service wishes to establish a weight value c beyond which there will be a surcharge. What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?

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  1. 7 July, 22:09
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    Answer: 23.7 lb

    Step-by-step explanation:

    Mean m = 15 lb

    Standard deviation d = 3.3 lb

    To determine the surcharge weight, we need to know the highest weight of 99% of the parcels.

    P (z
    Z = 2.33

    Since Z = (x - m) / d

    x = dZ + m

    x = 3.3*2.33 + 15

    x = 22.689 lb approximately

    x = 22.7 lb

    Therefore, the highest weight for 99% if the parcels is 22.7 lb.

    That is, the surcharge weight = 22.7 + 1 = 23.7 lb
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