What are partial products for 34×28
1 answer:
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Answer:
So the value of height that separates the bottom 10% of data from the top 90% is 30.3.
If the score is lower than 30.3 we consider this score as risk averse
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
Where and
We are interested in the bottom 10% of the data.
For this part we want to find a value a, such that we satisfy this condition:
(a)
(b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9
If we use condition (b) from previous we have this:
But we know which value of z satisfy the previous equation so then we can do this:
And if we solve for a we got
So the value of height that separates the bottom 10% of data from the top 90% is 30.3.
If the score is lower than 30.3 we consider this score as risk averse
Explanation:
When the points are plotted on a graph, it is easy to see that the slope of AC is -2 and the slope of BC is 1/2. These slope values have a product of -1, so the corresponding line segments are perpendicular to each other.
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If you have studied vectors, you can find the dot product of AC with BC:
AC = (-5, 3) -(-2, -3) = (-3, 6)
BC = (6, 1) -(-2, -3) = (8, 4)
The dot product is ...
(-3, 6)·(8, 4) = (-3)(8) + (6)(4) = -24+24 = 0
When the dot product of vectors is zero, they are perpendicular.
