Question

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?

Answer 1

`z =2.33`

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

`2.33= (X -17)/(3.7)`

`X= 17+ 2.33* 3.7 =25.62`

`C= 25.621 +1 = 26.621`

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:

`X \sim N(17,3.7)`

Where
`\mu=17` and
`\sigma=3.7`

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

`z=(x-\mu)/(\sigma)`

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:

`z =2.33`

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

`2.33= (X -17)/(3.7)`

`X= 17+ 2.33* 3.7 =25.62`

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

`C= 25.621 +1 = 26.621`