A random variable x x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a rand
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Like all problems that involve images within the question, we should definitely try to draw this out. In the picture above, I have done this.
Now, we can see that this is just a simple proportion problem. For every 2.5 cm of height of the flower, we are 2 cm from the opening, or aperture. For every 20 cm of height, how far are we? We can set up the problem like this:
20 ............2.5
-------- = ---------
...x ............. 2
where x is the unknown distance to the aperture from the flower. Now, we just need to get x by itself. A typical way of solving something like this is by doing "butterfly multiplication" which is really just a shortcut haha. Anyway, I can rewrite that equation ^ as:
20×2 = 2.5 × x
Then, to solve for x, we would divide both sides by 2.5. (If you don't know why that is, please let me know and I'll elaborate).
We would then have:
20×2
------- = x
2.5
Which then simplifies to:
x = 16
Try using the same logic for your second question, and if you get stuck, I'd be happy to help! please let me know if any of this doesn't make sense. :)
Now, we can see that this is just a simple proportion problem. For every 2.5 cm of height of the flower, we are 2 cm from the opening, or aperture. For every 20 cm of height, how far are we? We can set up the problem like this:
20 ............2.5
-------- = ---------
...x ............. 2
where x is the unknown distance to the aperture from the flower. Now, we just need to get x by itself. A typical way of solving something like this is by doing "butterfly multiplication" which is really just a shortcut haha. Anyway, I can rewrite that equation ^ as:
20×2 = 2.5 × x
Then, to solve for x, we would divide both sides by 2.5. (If you don't know why that is, please let me know and I'll elaborate).
We would then have:
20×2
------- = x
2.5
Which then simplifies to:
x = 16
Try using the same logic for your second question, and if you get stuck, I'd be happy to help! please let me know if any of this doesn't make sense. :)