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Solve Exercise 3. Dispersion measures for ungrouped data samples (C5 – Math 10 Kite)
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Solution of Exercises Lesson 1, page 41, Math Textbook 10 Kite, episode 2
In 5 long jumps, two friends Hung and Trung got results (unit: meters) of ., respectively
|
Hung |
2.4 |
2.6 |
2.4 |
2.5 |
2.6 |
|
Central |
2.4 |
2.5 |
2.5 |
2.5 |
2.6 |
a) Are your averages equal?
b) Calculate the variance of the sample statistics for each of the five long jumps. From there, it will tell you who has more stable long jump results.
Solution method
a) Average: \(\overline x = \frac{{{x_1} + {x_2} + … + {x_n}}}{n}\) , compare the results.
b) Variance:\({s^2} = \frac{1}{n}\left[ {{{\left( {{x_1} – \overline x } \right)}^2} + {{\left( {{x_2} – \overline x } \right)}^2} + … + {{\left( {{x_n} – \overline x } \right)}^2}} \right]\)
The smaller the variance, the more stable the results.
Solution guide
a) Your average result is equal: \(\overline {{x_H}} = \overline {{x_T}} = 2.5\) (m)
b) +) The sample variance of Hung and Trung’s statistics is:
\(s_H^2 = \frac{{{{\left( {2,4 – \overline {{x_H}} } \right)}^2} + {{\left( {2,6 – \overline {{) x_H}} } \right)}^2} + {{\left( {2,4 – \overline {{x_H}} } \right)}^2} + {{\left( {2,5 – \overline {{x_H}} } \right)}^2} + {{\left( {2,6 – \overline {{x_H}} } \right)}^2}}}{5} = 0.008\)
\(s_T^2 = \frac{{{{\left( {2,4 – \overline {{x_H}} } \right)}^2} + {{\left( {2,5 – \overline {{) x_H}} } \right)}^2} + {{\left( {2,5 – \overline {{x_H}} } \right)}^2} + {{\left( {2,5 – \overline {{x_H}} } \right)}^2} + {{\left( {2,6 – \overline {{x_H}} } \right)}^2}}}{5} = 0.004\)
+) 0.004 < 0.008, so we conclude: Trung's long jump result is stable.
Solution of Exercises Lesson 2, page 41, Math Textbook 10, Kite episode 2
The line graph in Figure 3 shows the GDP growth rate of Vietnam in the period 2012 – 2019.
a) Write a sample of the GDP growth rate statistics obtained from the graph in Figure 3.
b) Find the range of variation of that data sample.
c) Find the interquartile range of that data sample.
d) Calculate the variance and standard deviation of that data sample.
Solution method
a) Observe the chart
b) For data sample: \({x_1},{x_2},…,{x_n}\)
+) Step 1: Sort the data samples in non-decreasing order: \({X_1},{X_2},…,{X_n}\)
+) Step 2: Range: \(R = {X_n} – {X_1}\)
c) +) Step 1: Sort the data samples in non-decreasing order: \({X_1},{X_2},…,{X_n}\)
+) Step 2: \({Q_2} = {M_e} = \left\{ \begin{array}{l}{X_{k + 1}}\quad \quad \quad \quad \quad (n = 2k + 1)\\\frac{1}{2}({X_k} + {X_{k + 1}})\quad \;\,(n = 2k)\end{array} \right.\)
\({Q_1}\) is the median of the left half of the sorted data \({Q_2}\) (excluding \({Q_2}\) if n is odd)
\({Q_3}\) is the median of the right half sorted \({Q_2}\) (excluding \({Q_2}\) if n is odd)
Interquartile range: \({\Delta _Q} = {Q_3} – {Q_1}\)
d) +) Calculation of variance \({s^2} = \frac{1}{n}\left[ {{{\left( {{x_1} – \overline x } \right)}^2} + {{\left( {{x_2} – \overline x } \right)}^2} + … + {{\left( {{x_n} – \overline x } \right)}^2}} \right]\)
+) Standard Deviation \(s = \sqrt {{s^2}} \)
Solution guide
a) Based on the graph, we have the data sample as:
5.25 5.42 5.98 6.68 6.21 6.81 7.08 7.02
b) +) Sorting the data samples in non-decreasing order, we have:
5.25 5.42 5.98 6.21 6.68 6.81 7.02 7.08
+) The range of variation of that sample is: \(R = {x_{\max }} – {x_{\min }} = 7.08 – 5.25 = 1.83\)
c) +) Sorting the data samples in non-decreasing order, we have:
5.25 5.42 5.98 6.21 6.68 6.81 7.02 7.08
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+) The quartiles of the data sample are: \({Q_1} = 5.7,{Q_2} = 6.445,{Q_3} = 7.05\)
+) The interquartile range of the data sample is: \({Q_3} – {Q_1} = 1.35\)
d) +) The average GDP growth rate of Vietnam in the period 2012 – 2019 is:\(\overline x = \frac{{5,25{\rm{ + }}5.42{\rm{ + } }5,98{\rm{ + }}6,21{\rm{ + }}6.68\; + 6.81{\rm{ + }}7.02{\rm{ + }}7.08 }}{8} = 6,30625\) (%)
+) The variance of the data sample is: \({s^2} = \frac{{\left[ {{{\left( {5,25 – \overline x } \right)}^2} + {{\left( {5,42 – \overline x } \right)}^2} + … + {{\left( {7,08 – \overline x } \right)}^2}} \right]}}{8} \approx 0.44\)
+) The standard deviation of the data sample is: \(s = \sqrt {{s^2}} \approx 0.66\)(%)
Solve the exercises Lesson 3 page 41 Math textbook 10 Kite episode 2
The line chart in Figure 4 shows the selling price of gold during the first seven days of June 2021.
a) Write a sample of the gold selling price statistics obtained from the chart in Figure 4.
b) Find the range of variation of that data sample.
c) Find the interquartile range of that data sample.
d) Calculate the variance and standard deviation of that data sample.
Solution method
a) Observe the chart
b) For data sample: \({x_1},{x_2},…,{x_n}\)
+) Step 1: Sort the data samples in non-decreasing order: \({X_1},{X_2},…,{X_n}\)
+) Step 2: Range: \(R = {X_n} – {X_1}\)
c) +) Step 1: Sort the data samples in non-decreasing order: \({X_1},{X_2},…,{X_n}\)
+) Step 2: \({Q_2} = {M_e} = \left\{ \begin{array}{l}{X_{k + 1}}\quad \quad \quad \quad \quad (n = 2k + 1)\\\frac{1}{2}({X_k} + {X_{k + 1}})\quad \;\,(n = 2k)\end{array} \right.\)
\({Q_1}\) is the median of the left half of the sorted data \({Q_2}\) (excluding \({Q_2}\) if n is odd)
\({Q_3}\) is the median of the right half sorted \({Q_2}\) (excluding \({Q_2}\) if n is odd)
Interquartile range: \({\Delta _Q} = {Q_3} – {Q_1}\)
d) +) Calculation of variance \({s^2} = \frac{1}{n}\left[ {{{\left( {{x_1} – \overline x } \right)}^2} + {{\left( {{x_2} – \overline x } \right)}^2} + … + {{\left( {{x_n} – \overline x } \right)}^2}} \right]\)
+) Standard Deviation \(s = \sqrt {{s^2}} \)
Solution guide
a) Based on the graph, we have the data sample as:
5767 5757 5747 5737 5727 5747 5747 5722
b) The range of variation of that sample is: \(R = {x_{\max }} – {x_{\min }} = 5767 – 5722 = 45\)
c) +) Sorting the data samples in non-decreasing order, we have:
5722 5727 5737 5747 5747 5747 5757 5767
+) The quartiles of the data sample are: \({Q_1} = 5732,{Q_2} = 5747,{Q_3} = 5762\)
+) The interquartile range of the data sample is: \({Q_3} – {Q_1} = 30\)
d) +) The average gold price for the first 7 days of June 2021 is: \(\overline x = \frac{{5722{\rm{ + }}5727{\rm{ + }}5737{\rm { + }}5747{\rm{ + }}5747{\rm{ + }}5747{\rm{ + }}5757{\rm{ + }}5767}}{8} = 5743,875\) ( thousand VND/only)
+) The variance of the data sample is: \({s^2} = \frac{{\left[ {{{\left( {5722 – \overline x } \right)}^2} + {{\left( {5727 – \overline x } \right)}^2} + … + {{\left( {5767 – \overline x } \right)}^2}} \right]}}{8} \approx 193.35\)
+) The standard deviation of the data sample is: \(s = \sqrt {{s^2}} \approx 13.9\)( thousand VND/ only)
