Use repeated subtraction to divide.1÷1/5
1 answer:
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Answer:
So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.
Step-by-step explanation:
1) 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".
2) Solution to the problem
Let X the random variable that represent the scores on the LSAT of a population, and for this case we know the distribution for X is given by:
Where and
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 0.35 of the area on the left and 0.65 of the area on the right it's z=-0.385. On this case P(Z<-0.385)=0.35 and P(Z>-0.385)=0.65
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 35% of data from the top 65% (Or the 35 percentile) is 146.7.
Answer:
<em>Choose the first alternative</em>
Step-by-step explanation:
<u>Probabilities</u>
The requested probability can be computed as the ratio between the number of ways to choose two sophomores in alternate positions and the total number of possible choices
, i.e.
There are 6 sophomores and 14 freshmen to choose from each separate set. There are 20 students in total
We'll assume the positions of the selections are NOT significative, i.e. student A/student B is the same as student B/student A.
To choose 2 sophomores out of the 6 available, the first position has 6 elements to choose from, the second has now only 5
The total number of possible choices is
The probability is then
Choose the first alternative