The answer is 50
Let the number be x.
22 is 44%
x is 100%
Make a proportion:
22 : 44% = x : 100%
x = 22 * 100% : 44%
x = 50
Answer:
59
Step-by-step explanation:
Using the normal distribution, there is a 0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
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)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
Researching this problem on the internet, the parameters are given as follows:
![\mu = 150, \sigma = 9, n = 10, s = \frac{9}{\sqrt{10}} = 2.85](https://tex.z-dn.net/?f=%5Cmu%20%3D%20150%2C%20%5Csigma%20%3D%209%2C%20n%20%3D%2010%2C%20s%20%3D%20%5Cfrac%7B9%7D%7B%5Csqrt%7B10%7D%7D%20%3D%202.85)
The probability is <u>one subtracted by the p-value of Z when X = 157</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)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
Z = (157 - 150)/2.85
Z = 2.46
Z = 2.46 has a p-value of 0.993.
1 - 0.993 = 0.007.
0.007 = 0.7% probability that the mean score for 10 randomly selected people who took the LSAT would be above 157.
More can be learned about the normal distribution at brainly.com/question/15181104
#SPJ1
I'm going to assume you're asking where the point is located on a coordinate plane.
We have x = 0 and y = -4.
So, we start at (0, 0) and move down 4 units.
Best of Luck!