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:

a

explanation:

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )