Question 1205419
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I'll focus on the 1st problem only. I'll mark it in blue.
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The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 77 minutes and a standard deviation of 14 minutes. Suppose one person at the hot springs is randomly chosen. Let X = the amount of time that person spent at Grover Hot Springs . Round all answers to 4 decimal places where possible.


c. The park service is considering offering a discount for the 4% of their patrons who spend the least time at the hot springs. What is the longest amount of time a patron can spend at the hot springs and still receive the discount?
minutes.
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As mentioned by tutor Theo, you can use the online calculator at the link he posted. 
The calculator is very user friendly and provides a nice diagram. 


For exam purposes, your teacher may not let you use that calculator (or similar online calculators). It's possible your teacher may only allow something like a TI83 or TI84.  


If so, then check out this resource
<a href="https://www.statology.org/inverse-normal-distribution/">https://www.statology.org/inverse-normal-distribution/</a>
It explains how to reach the <font color=green>invNorm</font> function.


In this case you'll type in:
<font color=green>invNorm(0.04)</font>
*[illustration UploadedScreenshot_7.png]
The result of that command is approximately -1.750686


It will mean P(z < -1.750686) = 0.04 approximately.
The area under the curve to the left of z = -1.750686 is roughly 0.04
4% of the distribution is below the approximate cut off point of z = -1.750686


We'll use this to find the corresponding raw score x.
z = (x - mu)/sigma
z*sigma = x - mu
x = z*sigma + mu
x = -1.750686*14 + 77
x = 52.490396
x = <font color=red size=4>52.4904</font> when rounding to 4 decimal places.


The longest amount of time that can be spent is <font color=red size=4>approximately 52.4904 minutes</font>, and you will be in the "lowest 4%" category to get the discount. 
Any time higher than this will land you in the upper 96% of the distribution.


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If you aren't allowed a calculator, then use of a Z table is the only other option.


Here's one such table
<a href="https://www.ztable.net/">https://www.ztable.net/</a>
Tables like this should be found at the back of your stats textbook. 
Your teacher should likely hand them out as a reference sheet during exams.


The idea is to find the value of k such that P(Z < k) = 0.04
Unfortunately the exact value 0.04 is NOT inside the table.
The closest we can get is 0.04006
This value can be found at the row starting with "-1.7" and in the column with "0.05" at the top.
This means P(Z < -1.75) = 0.04006 approximately, which is roughly the same as P(Z < -1.75) = 0.04


Once you determine the proper z value, convert it to the raw score x as shown here:
z = (x - mu)/sigma
x = z*sigma + mu
x = -1.75*14 + 77
x = 52.5
We get a slightly different answer because we're using a less accurate z value.
Luckily 52.5 is close to 52.4904
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