Question 1195498: another big question. thank you. i wonder if i could solve these on desmos or wolfram alpha...
The mean number of miles (in thousands) that a car's tire will function before needing replacement is 68; the standard deviation is 16 (also in thousands). Assume a normal distribution.
a) If we test 44 tires, find the probability that the average miles (in thousands) before need of replacement is between 71.3 and 73.4. Round your answer to 4 decimal places.
P(71.3 < x bar < 73.4)=
b) Is the assumption that the distribution is normal necessary in order for you to answer part a?
Yes
No
c) If a randomly-selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 71.3 and 73.4. Round your answer to 4 decimal places.
P(71.3 < x < 73.4)=
d) Is the assumption that the distribution is normal necessary in order for you to answer part c?
Yes
No
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
I'm not too familiar with Desmos, but there may be a way to use that app with normal distribution applications.
Instead, I'll use WolframAlpha and GeoGebra.
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Part (a)
mu = 68 = population mean, in thousands
sigma = 16 = population standard deviation, in thousands
n = 44 = sample size
Compute the z score when xbar = 71.3
z = (xbar - mu)/( sigma/sqrt(n) )
z = (71.3 - 68)/( 16/sqrt(44) )
z = 1.368108 approximately
Repeat for xbar = 73.4
z = (xbar - mu)/( sigma/sqrt(n) )
z = (73.4 - 68)/( 16/sqrt(44) )
z = 2.238722 approximately
The task of computing P(71.3 < xbar < 73.4) is roughly equivalent to P(1.368108 < z < 2.238722)
If you were to type P(1.368108 < z < 2.238722) into WolframAlpha, exactly as shown, then it produces a result of approximately 0.0730521 as the screenshot below indicates.
I added the red arrow to help draw your attention to the result we're after.
Ignore the messy fraction they give you. It's not accurate because the input values were approximations. Even if the inputs were exact, we'd still should expect approximate results only. I'm not sure how Wolfram got that massive fraction. If I had to guess, the calculator truncated the approximate decimal value, and then built a fraction based on it.
Also, the plot they offer isn't correct either. Any normal distribution plot should have the ENTIRE curve above the x axis.
Edit: It appears the software coders at WolframAlpha have fixed the graphing glitch.
Luckily the 0.0730521 is correct. I used various other normal distribution calculators to verify this.
One such useful tool is GeoGebra.
The value 0.0730521 then rounds to 0.0731 when rounding to four decimal places.
Here's yet another tool to verify the answer
https://davidmlane.com/normal.html
This is a normal distribution calculator to compute areas under the curve.
Answer: 0.0731 (approximate)
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Part (b)
Answer: No
Reason:
The sample size n = 44 is larger than 30
By the central limit theorem, the xbar distribution is approximately normal even if the underlying parent distribution wasn't normal.
Therefore, we don't need to worry about if the original distribution is normal or not.
Further reading:
https://www.cuemath.com/data/central-limit-theorem/
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Part (c)
mu = 68
sigma = 16
We won't use the sample size this time since we're testing an individual tire (rather than a collection of them to get the sample mean)
Find the z score of each endpoint
z = (x - mu)/sigma
z = (71.3 - 68)/16
z = 0.20625
and
z = (x - mu)/sigma
z = (73.4 - 68)/16
z = 0.3375
Both decimal results are exact
The task of finding P(71.3 < x < 73.4) is exactly equivalent to P(0.20625 < z < 0.3375)
Type that second expression directly into WolframAlpha and you should get 0.0504278
Ignore the fractional result stated, and the plot is a bit off as well. We have the same weird issues mentioned in part (a). I verified that the decimal result is correct with the other calculators mentioned.
The value 0.0504278 rounds to 0.0504
Answer: 0.0504 (approximate)
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Part (d)
Answer: Yes
In contrast to part b, the assumption for normality is necessary because we used the Z distribution which is the normal distribution when mu = 0 and sigma = 1.
We cannot rely on the n > 30 condition because again we aren't using the sample size for part (c).
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