Answer:
99.1 is the 90th percentile of these scores.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 85
Standard Deviation, σ = 11
We are given that the distribution of score is a bell shaped distribution that is a normal distribution.
Formula:
![z_{score} = \displaystyle\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z_%7Bscore%7D%20%3D%20%5Cdisplaystyle%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
We have to find the value of x such that the probability is 0.90
Calculation the value from standard normal z table, we have,
![P(z \leq 1.282) = 0.90](https://tex.z-dn.net/?f=P%28z%20%5Cleq%201.282%29%20%3D%200.90)
Putting values, we get,
Thus, 99.1 is the 90th percentile of these scores.
I won't give you the answer because I am sure you are capable of finding it yourself once given a push, but what I would do is simplify all of the numbers into a mixed fraction (a b/c) and then go from there. Don't subtract the two numbers with the denominator of 25 first because they goes against PEMDAS. Change all of the fractions so they have like denominators, and once you solve it, just write the fraction part of the answer (b/c). If you have any other questions, just ask.
I would set the equations on top of each other and subtract
6x+y=21
- 3x-y=6
________
3x=15
Then solve for x
X=5
To solve for y, plug 5 in for x for one of the equations