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Exercise 1 Page 347
Calculate the measures of center and spread for each situation separately.
| Mean | Median | Range | IQR | Standard Deviation | |
|---|---|---|---|---|---|
| Without Make-up Test | 207.6 | 207.5 | 3 | 2 | 1.1 |
| With Make-up Test | 208 | 208 | 6 | 2 | 1.7 |
The following dot plot represents the scores of 10 students on a test.
Now, let's add the eleventh student. The one who took a make up test and made a score of 212.
We will find the mean, median, range, interquartile range, and standard deviation of the students before and after the 11^\text{th} student's score was added.
10 Students
First, we need to list the scores of the 10 students in order from least to greatest. 206,206,207,207,207,208,208,209,209,209
Mean
In order to find the mean μ, we find the sum of all values in the data set and divide it by the number of entries.
μ=206+206+207+207+207+208+208+209+209+209/10
AddTerms
Add terms
μ=2076/10
CalcQuot
Calculate quotient
μ=207.6
Thus, the mean score on the test was 207.6.
Median
In order to determine the median, we find the middle value or the mean of the two middle values of the ordered data. 206,206,207,207, 207, 208,208,209,209,209 Since there are 10 values, the median M is the mean of 5^\text{th} and 6^\text{th} values. Let's find it!
Range
The range R is the difference between the least and the greatest data values. 206,206,207,207, 207, 208,208,209,209, 209 The least data value is 206 and the greatest data value is 209. Let's find the range.
Thus, the range in the scores was 3.
Interquartile Range (IQR)
The IQR is the difference between the first quartile and the third quartile. The first quartile is the median of the lower half and the third quartile is the median of the upper half of the data set. 206,206, 207,207, 207, | 208,208, 209,209,209 Now, let's calculate the IQR!
As a result, the IQR is 2.
Standard Deviation
Standard deviation σ represents the average of the distance between individual data values and the mean. We can find the standard deviation using the following formula. σ=sqrt((x_1-μ)^2+(x_2-μ)^2+...+(x_(10)-μ)^2/n) Let's use a table in order to find the standard deviation step by step. Remember that we have already found the mean, μ= 207.6.
| Data Value, x | Deviation From Mean, (x-μ) | Squared Deviation, (x-μ)^2 |
|---|---|---|
| 206 | 206- 207.6=-1.6 | (-1.6)^2=2.56 |
| 206 | 206- 207.6=-1.6 | (-1.6)^2=2.56 |
| 207 | 207- 207.6=-0.6 | (-0.6)^2=0.36 |
| 207 | 207- 207.6=-0.6 | (-0.6)^2=0.36 |
| 207 | 207- 207.6=-0.6 | (-0.6)^2=0.36 |
| 208 | 208- 207.6=0.4 | (0.4)^2=0.16 |
| 208 | 208- 207.6=0.4 | (0.4)^2=0.16 |
| 209 | 209- 207.6=1.4 | (1.4)^2=1.96 |
| 209 | 209- 207.6=1.4 | (1.4)^2=1.96 |
| 209 | 209- 207.6=1.4 | (1.4)^2=1.96 |
Now, we can calculate the standard deviation.
σ=sqrt((x_1-μ)^2+(x_2-μ)^2+...+(x_(10)-μ)^2/n)
SubstituteValues
Substitute values
σ=sqrt(2.56+ 2.56+...+ 1.96/10)
AddTerms
Add terms
σ=sqrt(12.4/10)
CalcQuot
Calculate quotient
σ=sqrt(1.24)
CalcRoot
Calculate root
σ=1.11355
RoundDec
Round to 1 decimal place(s)
σ=1.1
Finally, we have found the standard deviation to be 1.1.
11 Students
Next, we will find the same measures of center and spread when the data includes the test score of the 11^(th) student. Let's add the score of 11^\text{th} student to the data set. 206,206,207,207,207,208,208,209,209,209,212
Mean
Again, we will start with the mean. Remember, the mean μ can be found by adding the scores and dividing by the number of scores.
μ=206+206+207+207+207+208+208+209+209+209+212/11
AddTerms
Add terms
μ=2288/11
CalcQuot
Calculate quotient
μ=208
The new mean of the scores is 208.
Median
This time we have 11 data points, so we only need to look for the middle one. 206,206,207,207,207, 208,208,209,209,209,212 With 11 values, the 6^\text{th} entry in the data set is the middle. Therefore, the new median is 208.
Range
Let's illustrate the least and the greatest data values. 206,206,207,207,207, 208,208,209,209,209, 212 Now, we can find the range.
The new range in scores is 6.
Interquartile Range (IQR)
Now, let's illustrate the first quartile and the third quartile. 206,206, 207,207,207, 208,208,209, 209,209,212 We can use these numbers to calculate the IQR!
As a result, the new IQR is 2.
Standard Deviation
We will find the standard deviation σ in the same way we did in Part A. Remember that, this time, the mean is 208.
| Data Value, x | Deviation From Mean, (x-μ) | Squared Deviation, (x-μ)^2 |
|---|---|---|
| 206 | 206- 208=-2 | (-2)^2=4 |
| 206 | 206- 208=-2 | (-2)^2=4 |
| 207 | 207- 208=-1 | (-1)^2=1 |
| 207 | 207- 208=-1 | (-1)^2=1 |
| 207 | 207- 208=-1 | (-1)^2=1 |
| 208 | 208- 208=0 | (0)^2=0 |
| 208 | 208- 208=0 | (0)^2=0 |
| 209 | 209- 208=1 | (1)^2=1 |
| 209 | 209- 208=1 | (1)^2=1 |
| 209 | 209- 208=1 | (1)^2=1 |
| 212 | 212- 208=4 | (4)^2=16 |
Now, we can calculate the standard deviation.
σ=sqrt((x_1-μ)^2+(x_2-μ)^2+...+(x_(11)-μ)^2/n)
SubstituteValues
Substitute values
σ=sqrt(4+ 4+...+ 16/11)
AddTerms
Add terms
σ=sqrt(30/11)
CalcQuot
Calculate quotient
σ=sqrt(2.72)
CalcRoot
Calculate root
σ=1.65144
RoundDec
Round to 1 decimal place(s)
σ=1.7
Finally, we have found that the new standard deviation is 1.7.
Table
Now, let's complete the given table.
| Mean | Median | Range | IQR | Standard Deviation | |
|---|---|---|---|---|---|
| Without Make-up Test | 207.6 | 207.5 | 3 | 2 | 1.1 |
| With Make-up Test | 208 | 208 | 6 | 2 | 1.7 |
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