Answer:
IQ scores of at least 130.81 are identified with the upper 2%.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X 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 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 100 and a standard deviation of 15.
This means that ![\mu = 100, \sigma = 15](https://tex.z-dn.net/?f=%5Cmu%20%3D%20100%2C%20%5Csigma%20%3D%2015)
What IQ score is identified with the upper 2%?
IQ scores of at least the 100 - 2 = 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![2.054 = \frac{X - 100}{15}](https://tex.z-dn.net/?f=2.054%20%3D%20%5Cfrac%7BX%20-%20100%7D%7B15%7D)
![X - 100 = 15*2.054](https://tex.z-dn.net/?f=X%20-%20100%20%3D%2015%2A2.054)
![X = 130.81](https://tex.z-dn.net/?f=X%20%3D%20130.81)
IQ scores of at least 130.81 are identified with the upper 2%.
Answer:multiplication Ex: 3 x 2 = 6
Division Ex: 6/2 =3
Step-by-step explanation:
3 x 2 = 3+3, add 3 2 times
6/2 = How many 2’s go into 6
Answer:
6 3/250
Step-by-step explanation:
So we see that we have 6 wholes and then 0.012 left over right?
Then we can also see that this would be pronounced as 12 "thousandths" and that would indicate 12/1000. Now we have a mixed fraction 6 12/1000
Simplifying this would get you
6 3/250.
Hope this helps :)