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.
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.
Best Answer
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 $\mu+a$.