Statistics: answer key for 3rd grade math exercises in PDF.

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The answer key to the math exercises in 3rd grade on statistics. Know how to calculate the mean, frequency and median of a statistical series in the third grade.

Exercise 1:

Let be the number of girls the number of boys is .

$\frac{11x+9,5\times \,(25-x)}{25}=10,4$

$11x+9,5\times \,(25-x)=10,4\times \,25$

$11x+9,5\times \,25-9,5x=260$

$x=\frac{22,5}{1,5}$

Conclusion: there are 15 girls and 10 boys in this class.

Exercise 2:

height in (m) 1.2 1.3 1.4 1.5 1.6
Number of employees 21 37 51 22 14

1. Calculate the average of this series.

$\overline{x}=\frac{1,2\times \,21+1,3\times \,37+1,4\times \,51+1,5\times \,22+1,6\times \,14\,}{21+37+51+22+14}$

$\overline{x}=\frac{200,1\,}{145}$

$\overline{x}=1,38$

The average height is 1.38 meters.

2. Determine the median of this series.

145 is an odd number, so look at the 73rd value.

The median is 1.4.

3. Interpret the results obtained previously.

If all the elements were the same size, each element would be 1.38 meters high.

There are as many elements that are higher than 1.4 meters as there are that are lower than 1.4 meters.

Exercise 3:

group of people
sizes in cm 130 145 155 160 170 175 180 190
workforce 3 5 11 25 36 20 8 2
cumulative workforce

1) calculate the average height of these people

$\overline{x}=\frac{130\times \,3+145\times 5+155\times 11+160\times 25+170\times 36+175\times 20+180\times 8+190\times 2}{3+5+11+25+36+20+8+2}$

$\overline{x}=\frac{390+725+1705+4000+3120+3500+1440+380}{110}$

$\overline{x}=\frac{15260}{110}$

$\overline{x}\simeq\,138,73$

2) Calculate the first quartile Q1, the median and the third quartile Q3 of this series.
For the 1st quartile:

The total number of employees is 110

$\frac{110}{4}=27,5$

So you have to look at the 28th value.

$Q_1=160\,\,cm$

For the median:

110 is even, so the 55th and 56th values must be averaged.

so the median is 170 cm .

For the 3rd quartile:

$\frac{3\times \,110}{4}=82,5$

So we have to look at the 83rd value.

$Q_3=175\,\,cm$

3) Calculate the range of the initial size series?

190-130=60 cm

Exercise 4:

10	20	30	40	50	60
15	25	35	45	55	65
240	628	556	370	356	190

1st line : Age noted “a
2nd line: Classroom Center
3rd line: Staffing

During the Christmas vacations, Mr. GLISSSE went skiing at L’alpe d’hurez. during one hour, the age of the people who went up in a telesiege was recorded (see the table)

1)Calculate the number of people who took the chairlift during this hour.

N=240+628+556+370+356+190=2,340 people

2) Complete the 2nd line of the table.

See the table

3)Calculate the average age of the skiers.

$\,\overline{x}=\frac{15\times \,240+25\times 628+35\times 556+45\times 370+55\times 356+65\times 190}{2340}$

$\,\overline{x}\simeq\,37,32$

Conclusion: the average age of the skiers is 37 years.

Exercise 5:

Here are 13 students’ scores on a math assignment:
8;9;19;17;6;18;18;8;14;12;9;10;11

1. Calculate the mean, rounded to the hundredth, of this series of scores.

$\overline{x}=\frac{8+9+19+17+6+18+18+8+14+12+9+10+11}{13}$

$\overline{x}=\frac{159}{13}$

${\color{DarkRed}\,\overline{x}\simeq\,12,23}$

2. calculate the percentage of students who have a grade above this class average.

grades above the class average are:19,17,18,14

There are 5 notes

$\frac{5}{13}\times \,100\simeq\,38,46\,%$

3. determine the median of this series of scores .

6;8;8;9;9;10;11;12;14;17;18;18;19

There are 13 notes, 13 is odd so the median is the 7th value.

The median is 11 .

Exercise 6:

$\overline{x}=\frac{2\times ,65+5\times ,72+4\times ,75+5\times ,80+3\times ,82+1\times ,90}{2+5+4+5+3+1}$

$\overline{x}=\frac{130+360+300+400+246+90}{20}$

$\overline{x}=\frac{1526}{20}$

$\overline{x}\simeq,76,3$

The average student pulse is 76.3

Exercise 7:

a) 230-29=201

The extent is 201 km.

$\overline{x}=\frac{195+165+195+.....+163+53+143,}{21}$

$\overline{x}=\frac{3554,}{21}$

$\overline{x}\simeq,169\,\,km$

If all stages were the same length, each stage would be 169 km long.

c) Let’s sort the values in ascending order:

29-53-143-154-157-158-163-165-166-168-174-182-182-195-195-195-197-210-216-222-230-

There are 28 values and 21 is an odd number

so the median is the 11th value or 174 km.

d) The extent is 53-29 = 24 km

The average is (53+29):2=82:2=41 km .

The median is also the mean.

Exercise 8:

1. $\overline{x}=\frac{0\times \,2+1\times 6+2\times 11+3\times 9+4\times 12+.....+19\times 11+20\times 5}{200}$

$\overline{x}=\frac{2068}{200}$

$\overline{x}=10,34$

Exercise 9:

1. $\overline{x}=\frac{18,6+19,4+20,8+15,9+17,7+21,1+19,8+15,2+17,2+16,5+20,5+21,9}{12}\simeq\,18,7$

Interpretation:

If the athlete had made all the same throws, he would have made throws of 18.7 meters.

2. $15,2<15,9<16,5<17,2<17,7<18,6<19,4<19,8<20,5<20,8<\,21,1<\,21,9$

There are 12 values and 12 is even, so let’s average the sixth and seventh values.

$\frac{18,6+19,4}{2}=19$

Interpretation:

There are as many throws below 19 meters as above.

Exercise 10:

There are two graders at the brevet des collèges:
the first has an average of 11 with 55 candidates and his colleague has only 9.5 with 45 candidates.
What is the overall average.

$\,\overline{x}=\frac{11\times \,55+9,5\times \,45}{55+45}$

$\,\overline{x}=\frac{1032,5}{100}$

$\,\overline{x}=10,325$

Exercise 11:
The gendarmes carried out a speed control on the side of a national road.

speed	[50;70[	[70;90[	[90;110[	[110;130[
workforce	15	90	35	5

Calculate the average speed of the motorists checked.

$\,\overline{v}=\frac{60\times \,15+80\times \,90+100\times \,35+120\times \,5}{15+90+35+5}$

$\,\overline{v}=\frac{12200}{145}$

$\,\overline{v}\simeq\,84,14$
Exercise 12:
French Polynesia has 219,500 inhabitants.

Their geographical distribution is represented by the following pie chart:

a. calculate the number of inhabitants of the Tuamotu-Gambier islands.

b. calculate the percentage of the population of the Leeward Islands in relation to the total population.

$\frac{26800}{219500}\times \,100=12,21%$

Exercise 13:

	France	Belgium	Italy	Spain	Other
number of wins	36	18	9	9	13
frequency (in percent)	$\frac{36}{85}\times \,100=42,35%$	$\frac{18}{85}\times \,100=21,18%$	$\frac{9}{85}\times \,100=18,59%$	$\frac{9}{85}\times \,100=18,59%$	$\frac{13}{85}\times \,100=15,3%$

Exercise 15:

This table gives the distribution of the 29,495,000 dwellings in France according to their category.

Category	Frequency (%)
Main residence	83,2
Vacant dwellings	6,8
Occasional housing	1
Secondary residences	9

1. Represent this frequency distribution by a pie chart.

Calculate the number of units in each category, rounding to the nearest thousand.

Category	Number of dwellings
Main residence	24 539 840
Vacant dwellings	2 005 660
Occasional housing	294 950
Secondary residences	2 654 550

Exercise 17:

a. The range for women’s wages is: 3470-1044=2426.

This means that the gap between the highest and lowest salary for women is 2,426 euros .

The range for men’s wages is: 2096-1002=1094.

This means that the gap between the highest and lowest salary for men is 1,094 euros.

The average salary for women is :

$\,\overline{x}=\frac{1090+1044+3470+1224+1250+1438+1072}{7}$

$\,\overline{x}=1512,57\,$ €

The average salary for men is :

$\,\overline{x}=\frac{1405+1070+1948+1525+1090+1002+1525+1968+1224+2096+1703+1126}{12}$

$\,\overline{x}=1473,5$ €

Exercise 21:

notes 6 7 9 10 11 12 14 15 16 19
workforce 3 4 4 2 1 3 2 4 1 2

1) There were 2+1+3+2+4+1+2=15 students who got the average out of 26 students.

$f=\frac{15}{26}=0,5769$ or 57.69%.

2) 26 is an even number, the median is the average of the 13th and 14th value, i.e. (11+12):2=11.5 .

3a) $\frac{26}{4}=6,5$ is the 7th value.

${\color{DarkRed}\,Q_1=7}$

$3\times \,6,5=19,5$ or the 20th value.

${\color{DarkRed}\,Q_3=15}$

3b) There are 3 + 4 + 4 + 2 + 1 + 3 = 17 students who scored below 12 .

$\frac{17}{26}\times \,100=65,38$ or about 65%.

Exercise 22:

One person stated that 25% of accidents occur between 8pm and midnight. Is this statement true? Justify the answer.
(4581 accidents occurred between 8:00 and 9:00 a.m.)

Numbers from 20 to 23h (there is no 24h in the table):

Between 20h and 23h

There are 5+3+3=11% of accidents that occur so the person is wrong.

Exercise 23:

“Here are the average monthly temperatures for two cities, in degrees Celsius:

MEXICO
J F M A M J
12,4 14,1 16,2 17,4 18,4 17,7
J A S O N D
16,7 16,8 16,3 15,1 13,9 12

BARCELONA
J F M A M J
9,5 10,3 12,4 14,6 17,7 21,5
J A S O N D
24,3 24,3 21,8 17,6 13,5 10,3

1- For each of these cities:
a-Calculate the range of the temperature series;

For Mexico City: 18.4- 12.4 = 6°C

For Barcelona: 24.3 – 9.5 = 14.8 °C

b- Estimate the average annual temperature ;

For mexico : $\overline{x}=\frac{12,4+14,1+16,2+17,4+18,4+17,7+16,7+16,8+16,3+15,1+13,9+12}{12}$

$\overline{x}=15,58^{\circ}C$

For Barcelona:

$\overline{x}=\frac{9,5+10,3+12,4+14,6+17,7+21,5+24,3+24,3+21,8+17,6+13,5+10,3}{12}$

$\overline{x}=16,48^{\circ}C$

Exercise 24:

Kevin tells his sister, “On average, we get paid more at KALU.”
What do you think about it?

Let’s calculate the average at HITI:

$\overline{x}=\frac{50\times \,168000+50\times \,120000}{50+50}$

$\overline{x}=\frac{8400000+6000000}{100}$

$\overline{x}=\frac{14400000}{100}$

$\overline{x}=144000$ €

Let’s calculate the average salary at KALU:

$\overline{x}=\frac{180000\times \,20+132000\times \,80}{20+80}$

$\overline{x}=\frac{3600000+10560000}{20+80}$

$\overline{x}=\frac{14160000}{100}$

$\overline{x}=141600$ €

Conclusion: employees are, on average, better paid at HITI so Kevin is wrong.

The answers to the math exercises in PDF on statistics in 3rd grade.

After having consulted the answers to these exercises on statistics in 3rd grade, you can return to the exercises in 3rd grade.

Third grade exercises.

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