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## Samples of size n = 90 are randomly selected from the population of numbers (0 through 9) produced by a random-number generator, and the variance is found for each sample. What is the distribution of the sample variances?

chi square distribution with df = n-1

Step-by-step explanation:

Given that sample of size n=90  are randomly selected from the population of numbers (0 through 9) produced by a random-number generator

Since sample size is large and randomness is followed we can assume that the variable follows a normal distribution.

Hence the sample variance would follow a chi square distribution with degree of freedom =

This is because we have is standard normal hence square will be a chisquare variate.  When we sum n variates we get chi square distribution with df n-1

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## In the data set, what is the variance? 6 8 1 9 4

Answer:  [B]:  “contains one point” .
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Explanation:
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Given:
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x + y  = 6  ;
x – y = 0 ;
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To solve for “x” ;

Consider the first equation:

x + y = 6 ;

subtract “y” from each side of the equation ; to isolate “x” on one side of the equation; and to solve for “x” ;

x + y – y = 6 – y ;

x = 6 – y ;
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Take the second equation:
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x – y = 0 ;

Solve for “x” ;

Add “y” to EACH SIDE of the equation; to isolate “x” on one side of the equation; and to solve for “x” ;

x – y + y = 0 + y ;
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x = y
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x = 6 – y

Substitute “x” for “y” ;

x = 6 – x ;

Add “x” to Each side of the equation:
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x + x = 6 – x + x ;

2x = 6 ;

Now, divide EACH SIDE of the equation by “2” ; to isolate “x” on one side of the equation; and to solve for “x” ;

2x/2 = 6/2 ;

x = 3 .
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Now, since “x = 3” ;  substitute “3” for “x” in both original equations; to see if we get the same value for “y” ;
_______________________________
x + y  = 6  ;
x – y = 0
________________________________
________________________________
x + y = 6 ;

3 + y = 6 ;

Subtract  “3” from each side of the equation; to isolate “y” on one side of the equation; and to solve for “y” ;

3 + y – 3 = 6 – 3 ;

y = 3 .
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Now, continue with the second equation; {Substitute “3” for “x” to see the value we get for “y”} ;
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The second equation given is:
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x – y = 0 ;

Substitute “3” for “x” to solve for “y” ;

3 – y = 0 ;

Subtract “3” from EACH side of the equation:

3 – y – 3 = 0 – 3 ;

-1y = -3  ;

Divide EACH side of the equation by “-1” ;  to isolate “y” on one side of the equation; and to solve for “y” ;

-1y/-1 = -3/-1 ;

y = 3 .
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So, for both equations, we have one value:  x = 3, y = 3;  or:  write as:
“(3, 3)” ;  { which is:  “one single point” ;  which is:  Answer choice:  [B] } .
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## It is desired to estimate the mean gpa of each undergraduate class at a large university. assume that the variance of the gpas is 1.44. how large a sample is necessary to estimate the mean gpa within 0.25 at the 99% confidence level

Mahtematical and statistical reasoning makes it evident that there is a mistake in the writing of the question and the digits were duplicated. So the right question is:

“You need to have a password with 5 letters followed by 3 odd digits between 0 and 9, inclusive. If the characters and digits cannot be used more than once, how many choices do you have for your passwor?”

The solution is:

5 letters from 26 with no repetition => 26*25*24*23*22 different choices.

odd digits are 1, 3, 5, 7 and 9 => 5 different digits to choose

3 digits from 5 with no repetition => 5*4*3 = 60

Then, the total number of choices is: 26*25*24*23*22*60 = 473,616,000

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## It has been reported that 63% of adults aged 65 and older got their flu shots last year. in a random sample of 300 adults aged 65 and over, find the mean, the variance, and standard deviation for the number who got their flu shots.

It has been reported that 63% of adults aged 65 and older got their flu shots last year. in a random sample of 300 adults aged 65 and over, find the mean, the variance, and standard deviation for the number who got their flu shots.

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## Find the​ mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n equals 123n=123​, p equals 0.85

False

Step-by-step explanation:

Suposse that we are given a function f(x) and a constant value h.

1. Case:

If we take the function g(x)=f(x)+h, then the graph of the function g(x) will be the graph of the funcion f(x) moved up or down.

2.Case:

If we take the function g(x)=hf(x), then the graph of the function g(x) will be the graph of the function f(x) just taller or shorter.

3.Case:

If we take the function g(x)=f(x-h), then the graph of the function g(x) will be the graph of the fuction f(x) moved horizontally.

4. Case:

If we take the function g(x)=f(hx), then the graph of the function g(x) will be tha graph of the function f(x) wither or thiner.

For example:

If we take f(x)=sin(x) and h=2. Then, if we take g(x)=sin(2x) then f(0)=g(0)=0, which means that the graph of the functiction is not moved up or down. However, f(π/2)=sin(π/2)=1 and g(π/2)=sin(π)=0 which gives us a hint that the graph of the function became thiner.

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## Brian collects a set of 20 data representing the lengths of worms he found in the garden. The variance of measurments is 36. What is the standard deviation from the mean?

Ur points are set up like this (x,y)….so ur x axis number, then ur y axis number…so 2 of the points that are on ur line are (0,12000) and (120,0)
u do not subtract 2000 by 4000…if u wanted to use those points, they would be (80,4000) and (100,2000)

slope (rate of change) = (y2 – y1) / (x2 – x1)
(0,12000) …x1 = 0 and y1 = 12000
(120,0) ……x2 = 120 and y2 = 0
slope = (0 – 12,000) / (120 – 0) = -12,000/120 = -100 <== ur slope

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## The following table shows the probability distribution for a discrete random variable. X 11 14 16 19 21 23 24 29 P(X) 0.07 0.21 0.17 0.25 0.05 0.04 0.13 0.08 The mean of the discrete random variable X is 18.59. What is the variance of X? Round your answer to the nearest hundredth.

Easy way is to list multipules of 6
6,12,18,24,30,36,42,48,54
so we can see that A doesn’t work because 20 doesn’t have 6 as a common factor
C works
D doesn’t work because of the 56

another way is to factor each
notice that 6=2*3
so factor each and see if we can find a 2*3 in each

A.
12=2*2*3, so there is a 6
20=2*2*5, not 6

B.
18=2*3*3, so there is a 6
24=2*2*3*3, so there is a 6
works

C.
30=2*3*5, there is a 6
54=2*3*3*3, there is a 6
works

D.
42=2*3*7, there is a 6
56=2*2*2*7, there isn’t a 6
nope

answes are B and C