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:
- 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:
The probability is <u>one subtracted by the p-value of Z when X = 157</u>, hence:
By the Central Limit Theorem
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
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In this expansion there are 5 terms:
3, x, 2, y and 6
And from all of these, the fifth one is 6
Hope it helped,
BioTeacher101
Total base premium = Summation of all base premiums
Base premiums are derived from insurance tables. For instance, 25/50 = $220 (body injury), 100 = $375 (property damage), At liability $250 = $185 and At comprehensive $250 = $102. Refer to figures attached.
That is,
Total base premium = 220+375+185+102 = $882
Answer:
x = -71
Step-by-step explanation:
If it correlates together than it goes down a total of -102.
He can make 27 different salads. Since each individual option has three different salads, and there are 9 options.
Basically multiply the number of total toppings by the number of toppings in each row
Answer: 27