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Lostsunrise [7]
3 years ago
14

For an F distribution, the number of degrees of freedom for the numerator a. must be larger than the number of degrees of freedo

m for the denominator. b. can be larger, smaller, or equal to the number of degrees of freedom for the denominator. c. must be smaller than the number of degrees of freedom for the denominator. d. must be equal to the number of degrees of freedom for the denominator.
Mathematics
1 answer:
Alecsey [184]3 years ago
7 0

Answer:

b. can be larger, smaller, or equal to the number of degrees of freedom for the denominator.

Step-by-step explanation:

The distribution of all possible values of the f statistic is called an F distribution with various degree of freedom. For an F distribution, the F statistic is greater than or equal to zero and as the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.

For an F distribution, the number of degrees of freedom for the numerator can be larger, smaller, or equal to the number of degrees of freedom for the denominator.

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Greatest common divisor of 24 ; 80 and 180 ?
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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
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2 years ago
The scores on the LSAT are approximately normal with mean of 150.7 and standard deviation of 10.2. (Source: www.lsat.org.) Queen
faltersainse [42]

Answer:

a=150.7 -0.385*10.2=146.773

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:

X \sim N(150.7,10.2)  

Where \mu=150.7 and \sigma=10.2

We want to find a value a, such that we satisfy this condition:

P(X>a)=0.65   (a)

P(X   (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:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

Z=-0.385

And if we solve for a we got

a=150.7 -0.385*10.2=146.773

So the value of height that separates the bottom 35% of data from the top 65% (Or the 35 percentile) is 146.7.  

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3 years ago
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