PLEASE HELP!!!
1 answer:
Remark
There are only 3 points. The 0.8 is a little bit skewed, but I would still use the mean.
Question 1
Mean = (x1 + x2 + x3)/3
Mean = (0.3 + 0.4 + 0.8) / 3
Mean = (1.5)/3 = 0.5
The balance point is 0.5
Question 2
It depends on what the data looks like.
See the graph below
With 3 points, it's pretty hard to tell. I calculated the middle one as the mean and it is what I take to be the central tendency, but do not be surprised if it is not. You could pick he left graph but how would you decide where the mode or median were?
Question 3
The measure of the center is the central tendency. It is what you use to figure out the dominant point of how the central tendency is found. There are not enough points to pick either the mode or the median.
There are only 3 points. The 0.8 is a little bit skewed, but I would still use the mean.
Question 1
Mean = (x1 + x2 + x3)/3
Mean = (0.3 + 0.4 + 0.8) / 3
Mean = (1.5)/3 = 0.5
The balance point is 0.5
Question 2
It depends on what the data looks like.
See the graph below
With 3 points, it's pretty hard to tell. I calculated the middle one as the mean and it is what I take to be the central tendency, but do not be surprised if it is not. You could pick he left graph but how would you decide where the mode or median were?
Question 3
The measure of the center is the central tendency. It is what you use to figure out the dominant point of how the central tendency is found. There are not enough points to pick either the mode or the median.
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Answer:
A. 16.385
B. 9.532
C. (16.358)^2 is different from (9.532)^2
a. sigma squared of non pregnant women is GREATER THAN sigma squared of pregnant women
Step-by-step explanation:
<h3>A and B. Standard Deviation 's' of both columns:</h3>The formula for the standard deviation is:
s = √(∑(x - μ)²/(n))
where,
∑ = is the sum function
x = a number from the set
μ = mean of the set
n = is the amount of the numbers in the set.
the column of non-pregnant women is
[73, 61, 104, 75, 85, 65, 62, 98, 92, 106]
- first find the mean of this set:
mean = μ = (sum of all the numbers)/(amount of numbers)
μ = [73 + 61 + 104 +75 + 85 + 65 + 62 + 98 + 92 + 106]/(10)
μ = 821/10
μ = 82.1
similarly, for pregnant women the mean is
μ = 80.125
- now, to find the standard deviation 's', first we need to find the variance 's²'. So, what you have to do is subtract each value in the column with the mean 'μ' and square the result.
for non-pregnant you will get:
[ 82.81, 445.21, 479.61, 50.41, 8.41, 292.41, 404.01, 252.81, 98.01, 571.21]
for pregnant you will get:
[ 66.015, 15.015, 97.515, 221.265, 199.51, 102.515, 1.265, 23.76]
- finally just sum all the numbers of a column and divide with the amount of numbers in that column (remember that col1 has 10 numbers and col2 has 8 numbers). And you will get your variance for both pregnant and non-pregnant women:
for non-pregnant = 2684.89/10
non-pregnant = 268.4 (this is the variance and it is denoted by s²)
for standard deviation 's', just take the square root of the variance
= 16.385.
similarly standard deviation of pregnant women can be found to be:
= 9.532.
A. non-pregnant 'S' = 16.385
B. pregnant 'S' = 9.532
<h3>C. CLAIM:</h3>you only have to show whether the variance of the above two columns are different or not.
And YES, the variances of the two column are indeed different, hence you make the CLAIM as written in question(C)
<h3>Multiple Choice:</h3>here you need to show how different are the two values:
recall the variances:
Column1 = 268.4 (non-pregnant)
Column2 = 9.532 (pregnant)
now you know that the variance of non-pregnant is GREATER THAN the variance of pregnant
yep exponential, since it close to graph of