Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
<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.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
In this problem:
- The mean is of 660, hence
.
- The standard deviation is of 90, hence
.
- A sample of 100 is taken, hence
.
The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

By the Central Limit Theorem



has a p-value of 0.8665.
1 - 0.8665 = 0.1335.
0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213
Juan did.
The line on his graph takes longer to get to 8 blocks.;-D
Answer:
x=2,x=7
Step-by-step explanation:
-3x^2+27x=42
x^2-9x+14=0
x^2-2x-7x+14=0
x*(xi2)-7(x-2)=0
(x-2)*(x-7)=0
x-2=0
x-7=0
x=2
x=7
You subtract 7% from 32=29.76
Then you subtract 8% from 29.76=27.379
(4+2)/(1+1)= 6/2= 3 is the slope
y + 2 = 3(x + 1)
y+2=3x+3
y= 3x+1