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
Answer:
Answer
5.0/5
23
Let's find the unit rate stemming from $18 for 4 games:
$18
------------- = $4.50/game
4 games
Let's do the same for $27 for 6 games:
$27
------- = $4.50/game (same as before)
6 g
Thus, the const. of prop. is $4.50/game, and the cost function is
C(x) = ($4.50/game)x, where x is the # of games played.
To begin with, we have to multiply 16 by 5, to determine how many ounces are in the bags. we get 80, no? we then divide 1680 by 80 to receive 21. your answer is 21.