Answer:
5
Step-by-step explanation:
1) add the parenthesis (5+4)
2) You get 9
3) you take your next number (-4) and because the signs are different you subtract and take the sign of the higher number (9).
Answer:
Step-by-step explanation:
By definition, the variance V of a population is defined as:
Where is the standard deviation
We know that , then we can solve the equation for the standard deviation
Finally the standard deviation is:
Sum of Interior Angles of any poligon, where n is the number of sides: =
(n-2) × 180°
---------------------
(7 - 2) * 180° =
5 * 180 =
900°
[d]
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:
- 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:
has a p-value of 0.841.
X = 406:
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:
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
Can you give me more info abt the problem plz so I can do the best I can to help