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The weight distribution of parcels sent in a certain manner is normal with mean of 12...

The weight distribution of parcels sent in a certain manner is normal with mean of 12 lb and standard deviation of 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 under the surcharge weight?

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Answer #1

Given that,

mean = \mu = 12

standard deviation = \sigma = 3

Using standard normal table,

P(Z < z) = 99%

= P(Z < z) = 0.99  

= P(Z < 2.33) = 0.99

z =2.33    Using standard normal z table,

Using z-score formula  

x= z * \sigma + \mu

x= 2.33 *3+12

x= 18.99

value of c =18.99

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