SOLUTION: 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 s
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Question 1205419: 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.
d. Find the Inter Quartile Range (IQR) for time spent at the hot springs.
Q1:
minutes
Q3:
minutes
IQR:
minutes
You can put this solution on YOUR website! 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.
you can use the z-score table or you can use a calculator.
a calculator is much easier.
to find the z-score that has 4% of the area under the normal distribution curve to the left of it, you use the value from an area option.
here are the results.
that z-score is -1.751.
to find the raw score, use the z-score formula.
z = (x-m)/s is the formula to use.
z-score = -1.751
m = population mean = 77
s = population standard deviation = 14
formula becomes -1.751 = (x - 77) / 14
solve for x to get x = -1.751 * 14 + 77 = 52.486.
that's the longest time you can get and still be in the bottom 4% of the distribution.
d. Find the Inter Quartile Range (IQR) for time spent at the hot springs.
to find the interquartile range, you want to get the area between Q1 and Q3.
you can find that by getting the area to the left of Q1 and to the left of Q3 and then subtracting the smaller area from the larger area to get the area in between.
you will get the z-score for Q1 and the z-score for Q3.
you will then solve for the raw score for Q1 and the raw score for Q3.
the minutes in the interquartile are the larger minutes minus the smaller minutes (minutes for Q3 minus minutes for Q1).
Q1 has 25% of the minutes below it.
the same calculator is used but you make the area .25.
select below and the calculator will tell you that the z-score is -.674.
with a mean of 77 and a standard deviation of 14, the z-score formula becomes -.674 = (x - 77) /14.
solve for x to get x = -.674 * 14 + 77 = 67.564.
that's the longest minutes you can have and still have a maximum of 25% of the area under the normal distribution curve to the left of it.
your answer is:
Q1:
67.564 minutes.
Q3 has 75% of the area under the normal distribution curve to the left of it.
the same calculator is used but you make the area .75.
select below and the calculator will tell you that the z-score is .674.
with a mean of 77 and a standard deviation of 14, the z-score formula becomes .674 = (x - 77) /14.
solve for x to get x = .674 * 14 + 77 = 86.436.
that's the longest minutes you can have and still have a maximum of 75% of the area under the normal distribution curve to the left of it.
your answer is:
Q3:
86.436 minutes
the interquartile range is the area between Q1 and Q3.
since you know the area below Q1 and the area below Q3, the area in between is the larger area minus the smaller area.
that's .75 minus .25 = .5
since you also know the minutes to the left of Q1 and the minutes to the left of Q1, you subtract Q1 minutes from Q3 minutes to get the minutes between Q1 and Q3 . Q3 minutes minus Q1 minutes is equal to 86.436 minus 67.564 = 18.872 minutes.
that's your inter-quartile range minutes.
your answer is:
IQR:
18.872 minutes.
here's the calculator results for finding the z-score with 25% of the area under the normal distribution curve to the left of it.
here's the calculator results for finding the z-score with 75% of the area under the normal distribution curve to the left of it.
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.
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.
In this case you'll type in:
invNorm(0.04)
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 = 52.4904 when rounding to 4 decimal places.
The longest amount of time that can be spent is approximately 52.4904 minutes, 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.
If you aren't allowed a calculator, then use of a Z table is the only other option.
Here's one such table https://www.ztable.net/
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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