C subtracted from the sum of x and y

find the mean the median the mode range and standard invitation of each data set that is obtained after adding the given content

timofeeve [1]

Answers:

Mean = 43.7

Median = 44.5

Mode = doesn't exist

Standard deviation = 4.78

Explanation:

First, we need to add the constant to each value, so the new data is:

33 + 11 = 44

38 + 11 = 49

29 + 11 = 40

35 + 11 = 46

27 + 11 = 38

34 + 11 = 45

36 + 11 = 47

28 + 11 = 39

41 + 11 = 52

26 + 11 = 37

Now, we can organize the data from least to greatest as:

37 38 39 40 44 45 46 47 49 52

Then, the mean is the sum of all the numbers divided by 10, because there are 10 values in the data. So, the mean is:

$\begin{gathered} \operatorname{mean}=\frac{37+38+39+40+44+45+46+47+49+52}{10} \\ \operatorname{mean}=43.7 \end{gathered}$

The median is the value that is located in the middle position of the organized data. Since there are 10 values, the values in the middle are the numbers 44 and 45, so the median can be calculated as:

$\operatorname{median}=\frac{44+45}{2}=44.5$

The mode is the value that appears more times in the data. Since all the values appear just one time, the mode doesn't exist.

To calculate the standard deviation, we will calculate first the variance.

The variance is the sum of the squared difference between each value and the mean, and then we divided by the number of values. So, the variance is equal to:

$\begin{gathered} (37-43.7)^2+(38-43.7)^2+(39-43.7)^2+(40-43.7)^2+ \\ (44-43.7)^2+(45-43.7)^2+(46-43.7)^2+(47-43.7)^2+ \\ (49-43.7)^2+(52-43.7)^2=228.1 \end{gathered}$ $\text{Variance}=\frac{228.1}{10}=22.81$

Finally, the standard deviation is the square root of the variance, so the standard deviation is: