The weight distribution of parcels sent in a certain manner is normal with mean value 17 lb and standard deviation 3.7 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?

Now we can use the z score formula in order to find the X value who satisfy the condition:

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

Where and

And the best way to solve this problem is using the normal standard distribution and the z score given by:

We can begin finding a value on the z table who accumulate 0.99 of the area on the left and 0.01 on the right and this value is:

Now we can use the z score formula in order to find the X value who satisfy the condition:

And we know that the value needs to have 1lb under the surchage weight so the final answer for this case would be:

answer:   rotate 270° clockwise about (7, 1) and reflect across the line x = 7

step-by-step explanation:

because i took test

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step-by-step explanation: