Answer:
I hope this helps a little bit
Here goes another question: 4/5%9/10 and please if you can with procces
2 answers:
When you divide a fraction by another fraction, it is the same thing as multiplying that fraction by its reciprocal.
The reciprocal is the 'flipped' version of a fraction where the numerator becomes the denominator and the denominator becomes the numerator.
for example, the reciprocal of 9/10 is 10/9
Knowing this information 4/5÷9/10 is the same as 4/5×10/9
Now we can solve...
Answer= 8/9
The reciprocal is the 'flipped' version of a fraction where the numerator becomes the denominator and the denominator becomes the numerator.
for example, the reciprocal of 9/10 is 10/9
Knowing this information 4/5÷9/10 is the same as 4/5×10/9
Now we can solve...
Answer= 8/9
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Answer:
And using a calculator, excel ir the normal standard table we have that:
And we can calculate the probability like this:Step-by-step explanation:
A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?
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 variable of interest of a population, and for this case we know the distribution for X is given by:
Where and
Since the distribution for X is normal then we know that the distribution for the sample mean is given by:
We can find the probability required like this:
And using a calculator, excel ir the normal standard table we have that:
And we can calculate the probability like this:
Answer:
Step-by-step explanation:
Recall that since X is uniformly distributed over the set [1,4] we have that the pdf of X is given by if
and 0 otherwise. In the same manner, the pdf of Y is given by
if
and 0 otherwise.
Note that if Y is in the interval (4,5] then Y>X by default. So, in this case we have that P(Y>X| y in (4,5]) = 1. We want to calculate the probability of having Y in that interval . That is
. Thus,
.
We want to proceed as follows. Using the total probability theorem, given two events A, B we have that
In this case, A is the event that Y>X and B is the event that Y is in the interval (4,5].
If we assume that X and Y are independent, then we have that the joint pdf of X,Y is given by when
. We can draw the region were Y>X and the function h(x,y) is different from 0. (The drawing is attached). This region is described as follows:
and
, then (the specifics of the calculations of the integrals are ommitted)
Thus,
