Which linear inequality will not have a shared solution set with the graphed linear inequality?
2 answers:
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In statistics, mean is the average in a set of data collected. To calculate the mean, you have to add up all the numbers in the table and divide it by the number of data that exists. For example, if the data set is 2 4 1 5 2 8, the mean will be:
mean = = 3.6
So, for this data set, the mean is 3.6.
The median is also a average number, however, the median is the middle value in the list. For example, with the data set above, the median is 3, because, as the set has an even number of data, the middle term would be the mean of the two middle number, in other words:
median = = 3
The standard deviation (σ) is the spread of data distribution, which means it's how "far" a number is from the mean of the set. The formula is
σ = √∑ (x - μ)²/N, where ∑ is the total sum; μ is the mean; and N is the number of data points in the sample. So, to calculate the standard deviation, using the example:
1) Calculate the mean of the set: μ=3.6
2) Find the difference between each data point and the mean and then, the square of each one:
(2-3.6)² = 2.56;
(4-3.6)² = 0.16
(1-3.6)² = 6.76
(5-3.6)² = 1.96
(2-3.6)² = 2.56
(8-3.6)² = 19.36
3) Add up all the squares: ∑= 33.36
4) Divide by the number of data points: =5.56
5) Take the square root: σ = 2.36
For the set of numbers above (2,4,1,5,2,8), the standard deviation is
σ = 2.36.
Answer:
The interquartile range of the data set is 5
The mean absolute deviation of the data set is 3.6
Step-by-step explanation:
* Lets explain how to find the interquartile range and the mean absolute
deviation (MAD)
- The steps to find the interquartile range is:
1- Arrange the values from the smallest to the largest
∴ The values are 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20
2- Find the median
- The median is the middle value after arrange them
* If there are two values in the middle take their average
∵ The values are 10 then the 5th and the 6th are the values
∵ The 5th is 12 and the 6th is 13
∴ The median =
∴ The median is 12.5
3- Calculate the median of the lower quartile
- The lower quartile is the median of the first half data values
∵ There are 10 values
∴ The first half is the first five values
∴ The first half values are 5 , 6 , 10 , 11 , 12
∵ The middle value is 10
∴ The median of lower quartile = 10
- Similar find the median of the upper quartile
- The upper quartile is the median of the second half data values
∵ There are 10 numbers
∴ The second half is the last five values
∴ The second half values are 13 , 14 , 15 , 18 , 20
∵ The middle value is 15
∴ The median of upper quartile = 15
4- The interquartile range (IQR) is the difference between the upper
and the lower medians
∴ The interquartile range = 15 - 10 = 5
* The interquartile range of the data set is 5
* Lets talk about the mean absolute deviation
- Mean absolute deviation (MAD) of a data set is the average distance
between each data value and the mean
- To find the mean absolute deviation of the data, start by finding
the mean of the data set.
1- Find the sum of the data values, and divide the sum by the
number of data values.
∵ The data set is 5 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 18 , 20
∵ Its sum = 5 + 6 + 10 + 11 + 12 + 13 + 14 + 15 + 18 + 20 = 124
∵ The mean = the sum of the data values/the number of the data
∵ The set has 10 numbers
∴ The mean = 124/10 = 12.4
2- Find the absolute value of the difference between each data value
and the mean ⇒ |data value – mean|
# I5 - 12.4I = 7.4
# I6 - 12.4I = 6.4
# I10 - 12.4I = 2.4
# I11 - 12.4I = 1.4
# I12 - 12.4I = 0.4
# I13 - 12.4I = 0.6
# I14 - 12.4I = 1.6
# I15 - 12.4I = 2.6
# I18 - 12.4I = 5.6
# I20 - 12.4I = 7.6
3- Find the sum of the absolute values of the differences.
∵ Their sum = 7.4 + 6.4 + 2.4 + 1.4 + 0.4 + 0.6 + 1.6 + 2.6 + 5.6 + 7.6
∴ Their sum = 36
4- Divide the sum of the absolute values of the differences by the
number of data values to find MAD
∴ MAD = The sum of the absolute values/number of the values
∵ The sum = 36
∵ The data set has 10 values
∴ MAD = 36/10 = 3.6
* The mean absolute deviation of the data set is 3.6