X + 388 = 388 + 152
x + 388 = 540
-388 -388
------------------------
x=152
Using the normal distribution, it is found that:
a) 68.2% of standardized test scores are between 406 and 644.
b) 31.8% of standardized test scores are less than 406 or greater than 644.
c) 2.3% of standardized test scores are greater than 763.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of 525, hence
.
- The standard deviation is of 119, hence
.
Item a:
The proportion is the <u>p-value of Z when X = 644 subtracted by the p-value of Z when X = 406</u>, hence:
X = 644:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{644 - 525}{119}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B644%20-%20525%7D%7B119%7D)
![Z = 1](https://tex.z-dn.net/?f=Z%20%3D%201)
has a p-value of 0.841.
X = 406:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{406 - 525}{119}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B406%20-%20525%7D%7B119%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
has a p-value of 0.159.
0.841 - 0.159 = 0.682.
0.682 = 68.2% of standardized test scores are between 406 and 644.
Item b:
Complementary event to the one found in item A, hence:
1 - 0.682 = 0.318.
0.318 = 31.8% of standardized test scores are less than 406 or greater than 644.
Item c:
The proportion is <u>1 subtracted by the p-value of Z when X = 763</u>, hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{763 - 525}{119}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B763%20-%20525%7D%7B119%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a p-value of 0.977.
1 - 0.977 = 0.023
0.023 = 2.3% of standardized test scores are greater than 763.
You can learn more about the normal distribution at brainly.com/question/24663213
Answer:
1
Step-by-step explanation:
70,001 > 70,000
Chow,...!
Answer:
Step-by-step explanation:
f(x)=2(x+4)(x-1)
=(2x+8)(x-1)
<h3>=2x²-2x+8x-8</h3>
=2x²+6x-8