The answer is
A. x=4
B. x=-4
A. x=4
B. x=-4
Use the formula M = NPQ to solve for the following variables.
2 answers:
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Answer:
Step-by-step explanation:
The area of a sector with measure and radius
is given by
.
What we're given:
of 12
of
Substituting given values, we get:
Answer:
a)
And we can find this probability using the normal standard table or excel:
b)
And we can find this probability with this difference:
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
c)
And we can find this probability using the complement rule and the normal standard table or excel:
d) We can consider unusual events values above or below 2 deviations from the mean
A value below 10.5 can be consider as unusual
A value abovr 31.7 can be consider as unusual
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".
Part a
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 on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability using the normal standard table or excel:
Part b
And we can find this probability with this difference:
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
Part c
And we can find this probability using the complement rule and the normal standard table or excel:
Part d
We can consider unusual events values above or below 2 deviations from the mean
A value below 10.5 can be consider as unusual
A value abovr 31.7 can be consider as unusual