A soccer player scores 3 goals in 2 games. How many goals are they expected to score in 9 games
1 answer:
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Answer:
Where and
We select a sample of n=16 and we are interested on the distribution of , since the distribution for X is normal then we can conclude that the distribution for
is also normal and given by:
Because by definition:
And for this case we have this:
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 IQ scores of a population, and for this case we know the distribution for X is given by:
Where and
We select a sample of n=16 and we are interested on the distribution of , since the distribution for X is normal then we can conclude that the distribution for
is also normal and given by:
Because by definition:
And for this case we have this:
Answer:
a)
b)
c)
d)
e)
f)
g)
h)
Step-by-step explanation:
(a) F0.05, 4, 7 (Round your answer to two decimal places.)
For this case we need a valueof the F distribution with 4 degrees of freedom for the numerator and 7 for the denominator that accumulates 0.05 of the area on the left tail. We can use the following excel code: "=F.INV(0.05,4,7)". And we got:
(b) F0.05, 7, 4 (Round your answer to two decimal places.)
For this case we need a valueof the F distribution with 7 degrees of freedom for the numerator and 4 for the denominator that accumulates 0.05 of the area on the left tail. We can use the following excel code: "=F.INV(0.05,7,4)". And we got:
(c) F0.95, 4, 7 (Round your answer to three decimal places.)
For this case we need a valueof the F distribution with 4 degrees of freedom for the numerator and 7 for the denominator that accumulates 0.95 of the area on the left tail. We can use the following excel code: "=F.INV(0.95,4,7)". And we got:
(d) F0.95, 7, 4 (Round your answer to three decimal places.)
For this case we need a valueof the F distribution with 7 degrees of freedom for the numerator and 4 for the denominator that accumulates 0.95 of the area on the left tail. We can use the following excel code: "=F.INV(0.95,7,4)". And we got:
(e) the 99th percentile of the F distribution with v1 = 8, v2 = 12 (Round your answer to two decimal places.)
So for this case we need a value on the F distribution with 8 degrees of freedom for the numerator and 12 for the denominator that accumulates 0.99 of the area on the left tail. And we can use the following excel code: "=F.INV(0.99,8,12)". And we got:
(f) the 1st percentile of the F distribution with v1 = 8, v2 = 12 (Round your answer to three decimal places.)
So for this case we need a value on the F distribution with 8 degrees of freedom for the numerator and 12 for the denominator that accumulates 0.01 of the area on the left tail. And we can use the following excel code: "=F.INV(0.01,8,12)". And we got:
(g) P(F ≤ 6.26) for v1 = 5, v2 = 4 (Round your answer to two decimal places.)
For this case we want to find the probability that the F distribution with 5 degrees on the numerator and 4 on the denominator would be less or equal than 6.26. We can use the following excel code: "=F.DIST(6.26,5,4,TRUE)". And we got
(h) P(0.177 ≤ F ≤ 4.74) for v1 = 10, v2 = 5 (Round your answer to two decimal places.)
For this case we want to find the probability that the F distribution with 10 degrees on the numerator and 5 on the denominator would be between 0.177 and 4.74. We can use the following excel code: "=F.DIST(4.74,10,5,TRUE)-F.DIST(0.177,10,5,TRUE)". And we got