Giáo Dục

Toán 10 Bài 3: Các số đặc trưng đo xu thế trung tâm của mẫu số liệu

Solve Math 10 Lesson 3: Feature numbers measure the central trend of the Creative Horizons data sample is an extremely useful document to help grade 10 students have more reference suggestions, easily compare results when doing math exercises from lessons 1 to 7 on pages 118, 119.

Solution for Math Textbook 10, Lesson 3, pages 118, 119 Creative horizons Volume 1 is compiled in detail, following the content in the textbook. Each problem is explained in detail. Thereby helping them consolidate and deepen the knowledge they have learned in the main program; can self-study, self-check their own learning results.

Math Solution 10: Typical numbers measure the central trend of the data sample

  • Math Solution 10 pages 118, 119 Creative horizons – Volume 1
    • Lesson 1 page 118
    • Lesson 2 page 118
    • Lesson 3 page 118
    • Lesson 4 page 118
    • Lesson 5 page 118
    • Lesson 6 page 119
    • Lesson 7 page 119

Math Solution 10 pages 118, 119 Creative horizons – Volume 1

Lesson 1 page 118

Find the mean, quartile, and mode of the following data samples:

a) 23; 41; 71; 29; 48; 45; 72; 41.

b) 12; 32; ninety three; 78; 24; twelfth; 54; 66; 78.

Suggested answer:

a) 23; 41; 71; 29; 48; 45; 72; 41.

+) Average number:overline x = frac{{23 + 41 + 71 + 29 + 48 + 45 + 72 + 41}}{8} = 46.25

+) Quartile:{Q_1},{Q_2},{Q_3}

Step 1: Sort the data samples in non-decreasing order:23;{rm{ }}29;{rm{ }}41;{rm{ }}41;;{rm{ }}45;{rm{ }}48;;71;72

Step 2: n = 8, is an even number, so {Q_2} = {M_e} = frac{1}{2}(41 + 45) = 43

{Q_1} is the median of half the data 23;{rm{ }}29;{rm{ }}41;{rm{ }}41. Hence {Q_2} = frac{1}{2}(29 + 41) = 35

{Q_3} is the median of half the data 45;{rm{ }}48;;71;72. Therefore {Q_3} = frac{1}{2}(48 + 71) = 59.5

+) Only the value 41 appears twice, more than the rest.

Hence the fad {M_o} = 41

b) 12; 32; ninety three; 78; 24; twelfth; 54; 66; 78.

+) Average number:overline x = frac{{12 + 32 + 93 + 78 + 24 + 12 + 54 + 66 + 78}}{9} approx 49.89

+) Quartile: {Q_1},{Q_2},{Q_3}

Step 1: Sort the data samples in non-decreasing order: 12;{rm{ }}12;{rm{ }}24;{rm{ }}32;{rm{ }}54;{rm{ }}66;{rm{ }}78; {rm{ }}78;;93

Step 2: n = 9, is an odd number, so{Q_2} = {M_e} = 54

{Q_1} is the median of half the data 12;{rm{ }}12;{rm{ }}24;{rm{ }}32. Hence {Q_2} = frac{1}{2}(12 + 24) = 18

{Q_3} is the median of half the data 66;{rm{ }}78;{rm{ }}78;;93. Hence {Q_3} = frac{1}{2}(78 + 78) = 78

+) The value 12 and the value 78 appear twice, more than the rest.

Hence the fad {M_o} = 12,{M_o} = 78.

Lesson 2 page 118

Find the mean, quartile, and mode of the following data samples:

The answer:

a) The data table is the frequency table.

The sample size is n = 6 + 8 + 10 + 6 + 4 + 3 = 37.

The sample mean is: overline x = frac{{23.6 + 25.8 + 28.10 + 31.6 + 33.4 + 37.3}}{{6 + 8 + 10 + 6 + 4 + 3}} approx 28.3

The value 28 has the highest frequency so the sample mode is Mo = 28.

Sorting the data samples in non-decreasing order, we get:

23; 23; 23; 23; 23; 23; 25; 25; 25; 25; 25; 25; 25; 25; 28; 28; 28; 28; 28; 28; 28; 28; 28; 28; thirty first; thirty first; thirty first; thirty first; thirty first; thirty first; 33; 33; 33; 33; 37; 37; 37.

Since the sample size is odd, the second quartile is Q .2 = 28.

The first quartile is the sample median: 23; 23; 23; 23; 23; 23; 25; 25; 25; 25; 25; 25; 25; 25; 28; 28; 28; 28. Hence Qfirst = 25.

The third quartile is the sample median: 28; 28; 28; 28; 28; thirty first; thirty first; thirty first; thirty first; thirty first; thirty first; 33; 33; 33; 33; 37; 37; 37. Hence Q3 = 31.

b) The data table is the relative frequency table.

The average is:overline x = frac{{0.0,6 + 2.0,2 + 4.0,1 + 5,0.1}}{{0.6 + 0.2 + 0.1 + 0.1}} = 1.3

Relative frequency is the ratio of frequency to sample size, so the value with the highest relative frequency has the largest frequency, so the value 0 has the largest frequency so the mode of the data sample is Mo = 0.

Assuming the sample size is n = 10, then:

The frequency of the value 0 is 0.6 . 10 = 6.

The frequency of the value 2 is 0.2 . 10 = 2.

The frequency of the value 4 is 0.1 . 10 = 1.

The frequency of the value 5 is 0.1 . 10 = 1.

Sorting the data in non-decreasing order, we get:

0; 0; 0; 0; 0; 0; 2; 2; 4; 5.

Since the sample size is even, the second quartile is Q .2 = 0.

The first quartile is the sample median: 0; 0; 0; 0; 0. Hence Qfirst = 0.

The third quartile is the sample median: 0; 2; 2; 4; 5. Hence Q3 = 2.

Lesson 3 page 118

An randomly picks 3 balls from a box containing many blue and red balls. An counts how many red balls out of the 3 taken out and returns the balls to the box. An repeated the test more than 100 times and recorded the results in the following table:

Number of red balls 0 first 2 3
Times ten 30 40 20

Find the mean, quartile, and mode of the table above.

Suggested answers

+) Average number: overline x = frac{{0.10 + 1.30 + 2.40 + 3.20}}{{100}} = 1.7

+) Quartile:{Q_1},{Q_2},{Q_3}

Step 1: Sort the data samples in non-decreasing order,underbrace {0,...,0}_{10},underbrace {1,...,1}_{30},underbrace {2,...,2}_{40},underbrace {3,...,3}_{20}.

Step 2: Since n = 100, is an even number, so {Q_2} = frac{1}{2}(2 + 2) = 2

{Q_1}is the median of half the data: underbrace {0,...,0}_{10},underbrace {1,...,1}_{30},underbrace {2,...,2}_{10}. Hence {Q_1} = frac{1}{2}(1 + 1) = 1

{Q_3} is the median of half the data underbrace {2,...,2}_{30},underbrace {3,...,3}_{20}. Hence {Q_3} = frac{1}{2}(2 + 2) = 2

+) Fashion {M_o} = 2

Lesson 4 page 118

In a contest, people record the time to complete a product of some experiments in the following table:

Time (unit: minutes)

5

6

7

8

35

Number of candidates

first

3

5

2

first

a) Find the mean, quartile and mode of the exam time of the above candidates.

b) Last year, the test time of candidates whose mean and median were both 7. Compare the overall exam time of candidates over the two years.

Suggested answer:

a.

+) Average number:overline x = frac{{1.5 + 3.6 + 5.7 + 2.8 + 1.35}}{{1 + 3 + 5 + 2 + 1}} = 9.08

+) Quartile:{Q_1},{Q_2},{Q_3}

Step 1: Sort the data samples in non-decreasing order, 5,6,6,6,7,7,7,7,7,8,8,35

Step 2: Since n = 12, is an even number, so {Q_2} = frac{1}{2}(7 + 7) = 7

{Q_1} is the median of half the data: 5,6,6,6,7 So {Q_1} = frac{1}{2}(6 + 6) = 6

{Q_3} is the median of half the data 7,7,7,8,8,35 So {Q_3} = frac{1}{2}(7 + 8) = 7.5

+) Fashion {M_o} = 7

b.

+) If the average number is compared: 9.08 > 7 so the overall exam time of the candidates this year is larger than the previous year.

+) If the median is compared: The median of two years is 7 so the overall test time of candidates in two years is the same.

Because there is one candidate whose test time is much longer than the other candidates => the comparison should be based on the median.

Lesson 5 page 118

Uncle Dung and Uncle Thu recorded the phone numbers that each person called every day for 10 days randomly selected from January 2021 in the following table:

Uncle Dung

2

7

3

6

first

4

first

4

5

first

Uncle Thu

first

3

first

2

3

4

first

2

20

2

a) Find the mean, quartile, and mode of each phone number that each caller uses the above numbers

b) On average, who has more phone calls?

c) If compared by median, who has more phone calls?

d) In your opinion, should the average or the median be used to compare who has more phone calls per day?

Suggested answers

a) Uncle Dung:

+) Average number: overline x = frac{{2 + 7 + 3 + 6 + 1 + 4 + 1 + 4 + 5 + 1}}{{10}} = 3.4

+) Quartile:{Q_1},{Q_2},{Q_3}

Step 1: Sort the data samples in non-decreasing order, 1,1,1,2,3,4,4,5,6,7

Step 2: Since n = 10, is an even number, so {Q_2} = frac{1}{2}(3 + 4) = 3.5

{Q_1} is the median of half the data: 1,1,1,2,3 So {Q_1} = 1

{Q_3} is the median of half of the data 4,4,5,6,7 So {Q_3} = 5

+) Fashion {M_o} = 1

Uncle Thu

+) Average number: overline x = frac{{1 + 3 + 1 + 2 + 3 + 4 + 1 + 2 + 20 + 2}}{{10}} = 3.9

+) Quartile: {Q_1},{Q_2},{Q_3}

Step 1: Sort the data samples in non-decreasing order, 1,1,1,2,2,2,3,3,4,20

Step 2: Since n = 10, is an even number, so {Q_2} = frac{1}{2}(2 + 2) = 2

{Q_1} is the median of half the data: 1,1,1,2,2 So {Q_1} = 1

{Q_3} is the median of half the data 2,3,3,4,20 So {Q_3} = 3

+) Fashion {M_o} = 1,{M_o} = 2

b) Since 3.9 > 3.4, according to the average number, Mr. Thu has more phone calls.

c) Since 3.5 > 2, according to the median, Mr. Dung has more phone calls.

d) Because in the data sample, there was one day that Mr. Thu had 20 phone calls, much larger than other days, so we should compare by median.

Lesson 6 page 119

The total number of points that the members of the International Mathematical Olympiad (IMO) team of Vietnam placed in 20 competitions are given in the following table:

Five

total score

Five

total score

Five

total score

Five

total score

2020

150

2015

151

2010

133

2005

143

2019

177

2014

157

2009

161

2004

196

2018

148

two thousand and thirteen

180

2008

159

2003

172

2017

155

2012

148

2007

168

2002

166

2016

151

2011

113

2006

131

2001

139

(Source: https://imo-office.org)

There is an opinion that the team’s test scores for the period 2001 – 2010 are higher than the period 2011 – 2020. Use the mean and median to check if the above opinion is correct.

Suggested answers

+) Period 2001 – 2010

Average number overline x = frac{{139 + 166 + 172 + 196 + 143 + 131 + 168 + 159 + 161 + 133}}{{10}} = 156.8

Sorting the data in non-decreasing order, we get: 131,133,139,143,159,161,166,168,172,196

Since n = 10 is even, the median is: {M_e} = frac{1}{2}(159 + 161) = 160

+) Period 2011 – 2020

Average number overline x = frac{{150 + 177 + 148 + 155 + 151 + 151 + 157 + 180 + 148 + 113}}{{10}} = 153

Sorting the data in non-decreasing order, we get:

113,;148,;148,;150,;151,;151,;155,;157,;177,;180

Since n = 10 is even, the median is: {M_e} = frac{1}{2}(151 + 151) = 151

+) Comparing by mean or median, we can see that the test scores of the 2001-2010 period are higher than the 2011-2020 period.

So the above statement is correct.

Lesson 7 page 119

The results of the midterm test for all students in grades 10A, 10B, and 10C are listed in the charts below.

a) Make a statistic of the number of students according to their scores in each class.

b) Compare the scores of students in those grades by mean, median, and mode.

Suggested answers

a)

Class 10A

The point

5

6

7

8

9

ten

HS No

first

4

5

8

14

8

Grade 10B

The point

5

6

7

8

9

ten

HS No

4

6

ten

ten

6

4

Grade 10C

The point

5

6

7

8

9

ten

HS No

first

3

17

11

6

2

b)

+) Class 10A

Average number overline x = frac{{5.1 + 6.4 + 7.5 + 8.8 + 9.14 + 10.8}}{{1 + 4 + 5 + 8 + 14 + 8}} = 8.35

Sorting the data in non-decreasing order, we get: 5,6,6,6,6,7,7,7,7,7,underbrace {8,...,8}_8,underbrace {9,...,9}_{14},underbrace {10,...,10}_8

Since n = 40, is an even number, the median is: {M_e} = frac{1}{2}(9 + 9) = 9

Fashion {M_e} = 9

+) Grade 10B

Average number overline x = frac{{5.4 + 6.6 + 7.10 + 8.10 + 9.6 + 10.4}}{{4 + 6 + 10 + 10 + 6 + 4}} = 7.5

Sorting the data in non-decreasing order, we get: 5,5,5,5,underbrace {6,..,6}_6,underbrace {7,...,7}_{10},underbrace {8,...,8}_{10 },underbrace {9,...,9}_6,10,10,10,10

Since n = 40, is an even number, the median is: {M_e} = frac{1}{2}(7 + 8) = 7.5

Fashion {M_e} = 7;{M_e} = 8.

+) Grade 10C

Average number overline x = frac{{5.1 + 6.3 + 7.17 + 8.11 + 9.6 + 10.2}}{{1 + 3 + 17 + 11 + 6 + 2}} = 7.6

Sorting the data in non-decreasing order, we get: 5,6,6,6,underbrace {7,...,7}_{17},underbrace {8,...,8}_{11},underbrace {9,...,9 }_6,10,10

Since n = 40, is an even number, the median is: {M_e} = frac{1}{2}(7 + 7) = 7

Fashion {M_e} = 7

+) Compare:

Average: 8.35 > 7.6 > 7.5 => The grades of students in descending order are 10A, 10C, 10B.

Median: 9 > 7.5 > 7=> The grades of students in descending order are 10A, 10B, 10C.

Fashion: Class 10A has 14 points, Class 10B has 10 points 7 and 10 points 8, Class 10C has 17 points 7. Therefore, comparing according to fashion, the scores of classes are reduced in order: 10A, 10B, 10C

Xem chi tiết bài viết

Toán 10 Bài 3: Các số đặc trưng đo xu thế trung tâm của mẫu số liệu

#Toán #Bài #Các #số #đặc #trưng #đo #thế #trung #tâm #của #mẫu #số #liệu


Tổng hợp: Hatienvenicevillas

Trả lời

Email của bạn sẽ không được hiển thị công khai.