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For a math assignment, I am required to find the range of data values that would include 95% of my data.

My data is the wait time for a specific bus. I have gathered 20 pieces of data for this, where each number represents how late(+)/early(-)/on time(0) the bus is.

This is the organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.

The link attached down below is my normal distribution graph where I have listed the mean (-0.55), standard deviation(2.4), and the graph information.

Normal Distribution Graph

What I think the range would be is (-5) - 4. I came to this conclusion by looking at the normal distribution graph above. Although I`m concerned about my answer.

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If this data is normally distributed with $\mu=-0.55$ and $\sigma=2.4$ we can say that X, the wait time for a specific bus, is distributed as follows: $$X\sim N(-0.55,(2.4)^2)$$ For 95% of the data to be included we want $P(\mu-a<X<\mu+a)=0.95$. As the normal distribution is symmetrical we can simplify this to finding: $P(X>\mu+a)=\frac{1-0.95}{2}=0.025$ or similarly $P(X<\mu-a)=0.025$. Using the Inverse Normal function (a website can be found here to do this) we find that $\mu-a=-5.2539$, such that: $$a=\mu+5.2539=-0.55+5.2539=4.7039$$ The range of data required is then $2\cdot a =2\cdot(4.7039)= 9.407$8 as the data ranges from $\mu-a$ to $\mu+a$.

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  • $\begingroup$ So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right... $\endgroup$ – Yashvi Shah Jan 4 at 23:49
  • $\begingroup$ Yes your initial estimate was quite accurate $\endgroup$ – Peter Foreman Jan 5 at 0:24
  • $\begingroup$ Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these? $\endgroup$ – Yashvi Shah Jan 5 at 1:14

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