SOLUTION: During the busy season, it is important for the shipping manager at ShipMundo to be able to estimate the time it takes the loading crew to load a truck. The shipping manager has fo

Algebra ->  Probability-and-statistics -> SOLUTION: During the busy season, it is important for the shipping manager at ShipMundo to be able to estimate the time it takes the loading crew to load a truck. The shipping manager has fo      Log On


   

Question 1201699: During the busy season, it is important for the shipping manager at ShipMundo to be able to estimate the time it takes the loading crew to load a truck. The shipping manager has found that she can model the load times using a normal distribution with a mean of 155 minutes and a standard deviation of 10 minutes.
Use this table or the ALEKS calculator to find the percentage of load times between 133 minutes and 159 minutes according to the model. For your intermediate computations, use four or more decimal places. Give your final answer to two decimal places (for example 98.23%).
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 64.15% (approximate)

=============================================================================

Work Shown

mu = 155 = population mean load time
sigma = 10 = population standard deviation of the loading times
x = load time in minutes

The task is to compute P(133 < x < 159) to find the percentage of load times between 133 minutes and 159 minutes.

Find the z score when x = 133
z = (x-mu)/sigma
z = (133-155)/10
z = -22/10
z = -2.20

Do the same for x = 159
z = (x-mu)/sigma
z = (159-155)/10
z = 4/10
z = 0.40

The task of finding P(133 < x < 159) is equivalent to P(-2.20 < z < 0.40)

I'll be using this Z table
https://www.ztable.net/
That table in the link should resemble the table your teacher has provided.
On that link, scroll down the page to see a few examples how to read that table.

Use such a table to find that
P(Z < -2.20) = 0.01390
P(Z < 0.40) = 0.65542
So,
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-2.20 < Z < 0.40) = P(Z < 0.40) - P(Z < -2.20)
P(-2.20 < Z < 0.40) = 0.65542 - 0.01390
P(-2.20 < Z < 0.40) = 0.64152
which leads back to
P(133 < x < 159) = 0.64152
This is the approximate probability of getting an x value between 133 and 159.

Convert that to a percentage
0.64152 --> 64.152%
That rounds to 64.15%

Approximately 64.15% of the load times are between 133 minutes and 159 minutes.

You could use a specialized stats calculator such as this one
https://davidmlane.com/normal.html
as an alternative to using a Z table.

This article goes over a few examples of how to calculate normal distribution probabilities on a TI84 calculator.
https://www.statology.org/normal-probabilities-ti-84-calculator/

A spreadsheet is another option.
Unfortunately I'm not familiar with the ALEKS calculator.