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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Answer:
10x²+28x+15
Step-by-step explanation:
(2x²+3)(4x+5)-8x(x²-2)=
8x³+12x+10x²+15-8x³+16x=
10x²+28x+15
Answer:
69 mins
Step-by-step explanation:
46/10=4.6 mins per wall
46/4=11.5 mins per wall
11.5 x 6=69