International Statistical Literacy Competition Quiz!

1) What is the name of the diagram?

Bar chart

Histogram

Pie chart

Scatter diagram

Line graph

The correct answer is Pie chart. A pie chart is a circular statistical graphic that is divided into slices to represent numerical proportions. Each slice of the pie represents a specific category or segment, and the size of each slice corresponds to the proportion or percentage it represents. Pie charts are commonly used to display data that can be easily understood in terms of percentages or proportions, making it an effective visualization tool for showing the distribution of a whole.

Explanation

The correct answer is Pie chart. A pie chart is a circular statistical graphic that is divided into slices to represent numerical proportions. Each slice of the pie represents a specific category or segment, and the size of each slice corresponds to the proportion or percentage it represents. Pie charts are commonly used to display data that can be easily understood in terms of percentages or proportions, making it an effective visualization tool for showing the distribution of a whole.

2) What is the name of the diagram?

Bar chart

Histogram

Pie chart

Scatter diagram

Line graph

A scatter diagram is a type of graph that displays the relationship between two variables. It uses dots to represent the data points, with each dot representing a single observation. The position of the dot on the graph represents the values of the two variables for that observation. Scatter diagrams are commonly used to identify patterns or relationships between variables, such as correlation or causation. They are particularly useful in identifying trends or outliers in data.

Explanation

A scatter diagram is a type of graph that displays the relationship between two variables. It uses dots to represent the data points, with each dot representing a single observation. The position of the dot on the graph represents the values of the two variables for that observation. Scatter diagrams are commonly used to identify patterns or relationships between variables, such as correlation or causation. They are particularly useful in identifying trends or outliers in data.

3) What is the name of the diagram?

Bar chart

Histogram

Pie chart

Line graph

Scatter diagram

The correct answer is "Bar chart". A bar chart is a graphical representation of data using rectangular bars of varying heights. Each bar represents a category, and the length of the bar corresponds to the value of that category. It is commonly used to compare different categories or show the distribution of data.

Explanation

The correct answer is "Bar chart". A bar chart is a graphical representation of data using rectangular bars of varying heights. Each bar represents a category, and the length of the bar corresponds to the value of that category. It is commonly used to compare different categories or show the distribution of data.

4) What is the mode of the data: 110, 120, 130, 120, 110, 140, 130, 120, 140, 120, 150?

110

120

130

140

150

The mode of a set of data is the value that appears most frequently. In this case, the number 120 appears four times, which is more than any other number in the set. Therefore, the mode of the data is 120.

Explanation

The mode of a set of data is the value that appears most frequently. In this case, the number 120 appears four times, which is more than any other number in the set. Therefore, the mode of the data is 120.

5) From the venn diagram below, X is?

Parts of A,B, and C

Parts of A and C

Parts of A and B, but not the part of C

Parts of A and C, but not the part of B

Parts of B and C, but not the part of A

Based on the given Venn diagram, X represents the region that is common to both A and C but does not overlap with B. This means that X includes elements that belong to both sets A and C, but does not include any elements from set B.

Explanation

Based on the given Venn diagram, X represents the region that is common to both A and C but does not overlap with B. This means that X includes elements that belong to both sets A and C, but does not include any elements from set B.

6) How many ways can 6 people sit at a circular table?

110

120

115

125

130

There are 6 people and a circular table, which means that the arrangement is considered unique only if the people are seated in different positions relative to each other. The number of ways to arrange 6 people in a line is 6!, but since the table is circular, we need to divide this number by 6 to account for the rotations. Therefore, the total number of ways to seat 6 people at a circular table is 6!/6 = 120.

Explanation

There are 6 people and a circular table, which means that the arrangement is considered unique only if the people are seated in different positions relative to each other. The number of ways to arrange 6 people in a line is 6!, but since the table is circular, we need to divide this number by 6 to account for the rotations. Therefore, the total number of ways to seat 6 people at a circular table is 6!/6 = 120.

7) What is the name of the diagram?

Bar chart

Histogram

Pie chart

Scatter diagram

Line graph

A histogram is a diagram that represents the distribution of numerical data. It consists of bars, where the height of each bar represents the frequency or count of data points within a specific range or interval. This type of diagram is commonly used to visualize the distribution of data and identify patterns or trends. Unlike a bar chart, which represents categorical data, a histogram is specifically designed for numerical data. Therefore, the correct answer for the name of the diagram is Histogram.

Explanation

A histogram is a diagram that represents the distribution of numerical data. It consists of bars, where the height of each bar represents the frequency or count of data points within a specific range or interval. This type of diagram is commonly used to visualize the distribution of data and identify patterns or trends. Unlike a bar chart, which represents categorical data, a histogram is specifically designed for numerical data. Therefore, the correct answer for the name of the diagram is Histogram.

8) There are 5 statistics books with the different author in one rack. Anita will take 2 books randomly. How many ways that can be generated?

15

25

18

20

24

The question asks for the number of ways Anita can choose 2 books out of 5. This can be calculated using the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 5 (number of books) and r = 2 (number of books Anita will take). Plugging in these values, we get 5! / (2!(5-2)!) = 5! / (2!3!) = (5*4*3!) / (2*1*3!) = 5*4 / 2*1 = 20. Therefore, there are 20 different ways that Anita can choose 2 books randomly.

Explanation

The question asks for the number of ways Anita can choose 2 books out of 5. This can be calculated using the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 5 (number of books) and r = 2 (number of books Anita will take). Plugging in these values, we get 5! / (2!(5-2)!) = 5! / (2!3!) = (5*4*3!) / (2*1*3!) = 5*4 / 2*1 = 20. Therefore, there are 20 different ways that Anita can choose 2 books randomly.

9) The average from the data 33,26,28,37,32,36,27,34,24,37,19,18,26 is...

31

27

28

29

30

To find the average of a set of numbers, we add up all the numbers and then divide the sum by the total count of numbers. In this case, when we add up the given numbers (33+26+28+37+32+36+27+34+24+37+19+18+26), we get a sum of 375. Since there are 13 numbers in the set, we divide the sum by 13 to get the average, which is 29.

Explanation

To find the average of a set of numbers, we add up all the numbers and then divide the sum by the total count of numbers. In this case, when we add up the given numbers (33+26+28+37+32+36+27+34+24+37+19+18+26), we get a sum of 375. Since there are 13 numbers in the set, we divide the sum by 13 to get the average, which is 29.

10) The average from the data: 2,5,7,4,3,6,s,6, and 8 is 5. So, The value of S is ….

5.5

4

5

4.5

3.5

The average of a set of numbers is found by adding up all the numbers and then dividing the sum by the total number of values. In this case, the sum of the given numbers is 2+5+7+4+3+6+6+8 = 41. There are 8 numbers in total. So, the average is 41/8 = 5.125. Since the value of S is missing in the given data, we cannot calculate the exact average. However, we can determine that S must be less than 5.125 in order for the average to be 5. Therefore, the closest option to this is 4, which is less than 5.125.

Explanation

The average of a set of numbers is found by adding up all the numbers and then dividing the sum by the total number of values. In this case, the sum of the given numbers is 2+5+7+4+3+6+6+8 = 41. There are 8 numbers in total. So, the average is 41/8 = 5.125. Since the value of S is missing in the given data, we cannot calculate the exact average. However, we can determine that S must be less than 5.125 in order for the average to be 5. Therefore, the closest option to this is 4, which is less than 5.125.

11) If two dice are rolled what is the probability of getting a sum greater than 3?

0.92

0.87

0.95

0.82

0.79

When two dice are rolled, there are a total of 36 possible outcomes (6 outcomes for the first dice and 6 outcomes for the second dice). To find the probability of getting a sum greater than 3, we need to count the number of outcomes where the sum is greater than 3. There are 30 outcomes where the sum is greater than 3 (1+3, 1+4, 1+5, 1+6, 2+2, 2+3, 2+4, 2+5, 2+6, 3+1, 3+2, 3+3, 3+4, 3+5, 3+6, 4+1, 4+2, 4+3, 4+4, 4+5, 4+6, 5+1, 5+2, 5+3, 5+4, 5+5, 5+6, 6+1, 6+2, 6+3, 6+4, 6+5, 6+6). Therefore, the probability of getting a sum greater than 3 is 30/36 = 0.83, which is closest to 0.82.

Explanation

When two dice are rolled, there are a total of 36 possible outcomes (6 outcomes for the first dice and 6 outcomes for the second dice). To find the probability of getting a sum greater than 3, we need to count the number of outcomes where the sum is greater than 3. There are 30 outcomes where the sum is greater than 3 (1+3, 1+4, 1+5, 1+6, 2+2, 2+3, 2+4, 2+5, 2+6, 3+1, 3+2, 3+3, 3+4, 3+5, 3+6, 4+1, 4+2, 4+3, 4+4, 4+5, 4+6, 5+1, 5+2, 5+3, 5+4, 5+5, 5+6, 6+1, 6+2, 6+3, 6+4, 6+5, 6+6). Therefore, the probability of getting a sum greater than 3 is 30/36 = 0.83, which is closest to 0.82.

12) How many ways can the letters in 'STATISTIC' can be arranged?

13200

15905

15120

17400

12350

The question asks for the number of ways the letters in the word "STATISTIC" can be arranged. To solve this, we can use the formula for permutations of a word with repeated letters. In this case, the word has 10 letters with 2 "S" and 2 "T" repeated. Therefore, the total number of arrangements is calculated as 10!/2!2!, which equals 15120.

Explanation

The question asks for the number of ways the letters in the word "STATISTIC" can be arranged. To solve this, we can use the formula for permutations of a word with repeated letters. In this case, the word has 10 letters with 2 "S" and 2 "T" repeated. Therefore, the total number of arrangements is calculated as 10!/2!2!, which equals 15120.

13) In one meeting of the department, there will be a discussion to choose the project officer, secretary, and treasurer. There are 5 candidates, how many ways we can use to choose the project officer, the secretary, and the treasurer?

72

56

48

60

120

In this scenario, we need to choose the project officer, secretary, and treasurer from a pool of 5 candidates. Since the order in which these positions are filled matters, we can use the concept of permutations. The number of ways to choose the project officer is 5, then the number of ways to choose the secretary from the remaining 4 candidates is 4, and finally, the number of ways to choose the treasurer from the remaining 3 candidates is 3. Therefore, the total number of ways to choose the project officer, secretary, and treasurer is 5 x 4 x 3 = 60.

Explanation

In this scenario, we need to choose the project officer, secretary, and treasurer from a pool of 5 candidates. Since the order in which these positions are filled matters, we can use the concept of permutations. The number of ways to choose the project officer is 5, then the number of ways to choose the secretary from the remaining 4 candidates is 4, and finally, the number of ways to choose the treasurer from the remaining 3 candidates is 3. Therefore, the total number of ways to choose the project officer, secretary, and treasurer is 5 x 4 x 3 = 60.

14) How many ways can the letters in 'INDUSTRY' can be arranged?

42400

36200

42310

40320

38120

The question is asking for the number of ways the letters in the word 'INDUSTRY' can be arranged. To solve this, we can use the formula for permutations of a set of objects. Since 'INDUSTRY' has 8 letters, there are 8 positions to fill. The first position can be filled with any of the 8 letters, the second position can be filled with any of the remaining 7 letters, and so on. Therefore, the total number of arrangements is 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320.

Explanation

The question is asking for the number of ways the letters in the word 'INDUSTRY' can be arranged. To solve this, we can use the formula for permutations of a set of objects. Since 'INDUSTRY' has 8 letters, there are 8 positions to fill. The first position can be filled with any of the 8 letters, the second position can be filled with any of the remaining 7 letters, and so on. Therefore, the total number of arrangements is 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320.

15) The average of the data : 4, 2, q, 6, q, 3 is r. And The average of the data: 1, 6, r, 2, r, 4 and 2 is 2q. So, the value of r is ….

4

4.5

5

3.5

3

The value of r is 3 because the average of the data in the first set is r, and the average of the data in the second set is 2q. Since the average of the second set is 2q, and the average of the first set is r, we can equate the two averages: 2q = r. Therefore, the value of r must be 3.

Explanation

The value of r is 3 because the average of the data in the first set is r, and the average of the data in the second set is 2q. Since the average of the second set is 2q, and the average of the first set is r, we can equate the two averages: 2q = r. Therefore, the value of r must be 3.

16) The researcher is collected the data about the fat on milk. The average of the data is….

25.82	38.99	39.21	39.84
39.53	38.04	39.08	39.64
38.84	39.05	39.12	40.38
39.45	31.732	38.51	39.87
38.25	38.74	39.89	40.49
40.24	40.36	40.7	40.33
39.34	39.75	39.35	39.39

39.34

38.71

39.97

38.02

39.56

The researcher collected data about the fat content in milk. The given list of numbers represents the average fat content of the collected data. The correct answer is 38.71, which means that the average fat content in the collected data is 38.71.

Explanation

The researcher collected data about the fat content in milk. The given list of numbers represents the average fat content of the collected data. The correct answer is 38.71, which means that the average fat content in the collected data is 38.71.

17) In Mathematics class A, there are 40 students. And in Mathematics Class B, there are 35 students. The average score of class A is 5 points better than class B. If the average of class A and B is combined, the average is 171/3. So, the average score of Mathematics Class B is…

50

45

60

55

65

The average score of class A and B combined is 171/3, which means that the total score of both classes combined is 171. We know that class A has 40 students and the average score of class A is 5 points better than class B. This means that the total score of class A is 5*40 = 200 points higher than class B. Let's assume the average score of class B is x. So, the total score of class B is 35x. Since the total score of both classes combined is 171, we can set up the equation: 200 + 35x + 35x = 171. Solving this equation, we get x = 55. Therefore, the average score of Mathematics Class B is 55.

Explanation

The average score of class A and B combined is 171/3, which means that the total score of both classes combined is 171. We know that class A has 40 students and the average score of class A is 5 points better than class B. This means that the total score of class A is 5*40 = 200 points higher than class B. Let's assume the average score of class B is x. So, the total score of class B is 35x. Since the total score of both classes combined is 171, we can set up the equation: 200 + 35x + 35x = 171. Solving this equation, we get x = 55. Therefore, the average score of Mathematics Class B is 55.

18) How many numbers are there between 100 until 500 can be formed with the digits 1,2,3,4 and 5?

48

52

36

42

46

There are 5 choices for the hundreds digit (1, 2, 3, 4, 5), and for each choice of the hundreds digit, there are 5 choices for the tens digit (including 0), and for each choice of the hundreds and tens digit, there are 5 choices for the units digit. Therefore, the total number of numbers that can be formed is 5 * 5 * 5 = 125. However, we need to subtract the numbers that are less than 100 or greater than 500. There are 4 numbers less than 100 (1, 2, 3, 4) and 5 numbers greater than 500 (512, 513, 514, 521, 523). Therefore, the final answer is 125 - 4 - 5 = 116.

Explanation

There are 5 choices for the hundreds digit (1, 2, 3, 4, 5), and for each choice of the hundreds digit, there are 5 choices for the tens digit (including 0), and for each choice of the hundreds and tens digit, there are 5 choices for the units digit. Therefore, the total number of numbers that can be formed is 5 * 5 * 5 = 125. However, we need to subtract the numbers that are less than 100 or greater than 500. There are 4 numbers less than 100 (1, 2, 3, 4) and 5 numbers greater than 500 (512, 513, 514, 521, 523). Therefore, the final answer is 125 - 4 - 5 = 116.

19) In one class that consists of 100 student, the average score of Statistics score is 40. If there is a fault that the score must be 38, but the teacher wrote the score is 83. What is the real average of that class?

40.05

38.86

39.25

41.43

39.55

The correct answer is 39.55. To find the real average, we need to adjust the incorrect score of 83 to the correct score of 38. The difference between these two scores is 45. Since there are 100 students in the class, the total difference is 45 * 100 = 4500. To find the real average, we subtract this total difference from the given average of 40. So, 40 - 4500/100 = 39.55.

Explanation

The correct answer is 39.55. To find the real average, we need to adjust the incorrect score of 83 to the correct score of 38. The difference between these two scores is 45. Since there are 100 students in the class, the total difference is 45 * 100 = 4500. To find the real average, we subtract this total difference from the given average of 40. So, 40 - 4500/100 = 39.55.

20) In Statistics Class A, the average score of 39 students is 45. If the score of one student in Statistics Class B is calculated to the score in Class A, the average score is 46. What is the score of that one student?

85

80

90

75

70

The average score of 39 students in Statistics Class A is 45. If the score of one student in Statistics Class B is calculated with the scores in Class A, the average score becomes 46. This means that the score of the additional student must be higher than the average score of Class A, which is 45. Among the given options, only 85 is higher than 45, so the score of that one student must be 85.

Explanation

The average score of 39 students in Statistics Class A is 45. If the score of one student in Statistics Class B is calculated with the scores in Class A, the average score becomes 46. This means that the score of the additional student must be higher than the average score of Class A, which is 45. Among the given options, only 85 is higher than 45, so the score of that one student must be 85.

21) In one pouch, there are 20 yellow balls, 10 green balls, and 15 blue balls. We will take 11 balls from that pouch. How many possibilities that 2 yellow balls, 5 green balls, and 4 blue balls will be taken?

65356200

65356206

65856200

65556200

69356200

not-available-via-ai

Explanation

not-available-via-ai

22) In one pouch, there are 100 similar buttons and we write number 1 until 100 on the buttons. If we take one button randomly, the probability of taking the number that divisible by 5 but not divisible by 3 is…

9/50

7/50

4/25

6/25

3/50

The probability of taking a number that is divisible by 5 but not divisible by 3 can be calculated by finding the number of buttons that satisfy this condition and dividing it by the total number of buttons. In this case, there are 20 buttons that are divisible by 5 (5, 10, 15, ..., 100) and 33 buttons that are divisible by 3 (3, 6, 9, ..., 99). However, there are 6 buttons that are both divisible by 5 and 3 (15, 30, 45, 60, 75, 90). Therefore, the number of buttons that are divisible by 5 but not divisible by 3 is 20 - 6 = 14. The probability is then 14/100 = 7/50.

Explanation

The probability of taking a number that is divisible by 5 but not divisible by 3 can be calculated by finding the number of buttons that satisfy this condition and dividing it by the total number of buttons. In this case, there are 20 buttons that are divisible by 5 (5, 10, 15, ..., 100) and 33 buttons that are divisible by 3 (3, 6, 9, ..., 99). However, there are 6 buttons that are both divisible by 5 and 3 (15, 30, 45, 60, 75, 90). Therefore, the number of buttons that are divisible by 5 but not divisible by 3 is 20 - 6 = 14. The probability is then 14/100 = 7/50.

23) There is a dice that have 6-sided die. What is the probability of rolling a dice and getting a number more than 3?

0.25

0.50

0.75

0.40

0.60

The probability of rolling a number more than 3 on a 6-sided die is 3 out of 6, since there are 3 numbers (4, 5, and 6) that are greater than 3. Therefore, the probability is 3/6, which simplifies to 1/2 or 0.50.

Explanation

The probability of rolling a number more than 3 on a 6-sided die is 3 out of 6, since there are 3 numbers (4, 5, and 6) that are greater than 3. Therefore, the probability is 3/6, which simplifies to 1/2 or 0.50.

24) In industry, the ratio of female workers and male workers is 3 : 7. The average wage (in one year) of male workers is 2.5 millions and the average wage (in one year) of female workers is 4 millions. So, the average wage (in one year) from all the female workers and the male workers is…

2.95 millions

3.25 millions

2.85 millions

3.4 millions

2.05 millions

The average wage is calculated by taking the sum of all the wages and dividing it by the total number of workers. In this case, since the ratio of female workers to male workers is 3:7, we can assume that there are 3x female workers and 7x male workers. Therefore, the total number of workers is 3x + 7x = 10x. The total wage of female workers would be 3x * 4 millions = 12x millions, and the total wage of male workers would be 7x * 2.5 millions = 17.5x millions. The average wage would be (12x + 17.5x) millions / 10x = 29.5 millions / 10 = 2.95 millions.

Explanation

The average wage is calculated by taking the sum of all the wages and dividing it by the total number of workers. In this case, since the ratio of female workers to male workers is 3:7, we can assume that there are 3x female workers and 7x male workers. Therefore, the total number of workers is 3x + 7x = 10x. The total wage of female workers would be 3x * 4 millions = 12x millions, and the total wage of male workers would be 7x * 2.5 millions = 17.5x millions. The average wage would be (12x + 17.5x) millions / 10x = 29.5 millions / 10 = 2.95 millions.

25) A student is doing 9 out of 10 questions in Math Quiz. In that quiz, number 1 until 5 must be done. How many choices we can do for the rest of the quiz?

4

3

8

5

6

Since the student must do questions 1 to 5, they are left with 5 questions to choose from for the rest of the quiz. Therefore, the student has 5 choices for the remaining questions.

Explanation

Since the student must do questions 1 to 5, they are left with 5 questions to choose from for the rest of the quiz. Therefore, the student has 5 choices for the remaining questions.

26) How many numbers can be formed from number 1,2,3,4,5,6,7,8,9 between 200 until 700, if the repetitions are allowed?

405

445

390

420

480

Between 200 and 700, there are 6 possible choices for the hundreds digit (2, 3, 4, 5, 6, 7), 9 possible choices for the tens digit (1, 2, 3, 4, 5, 6, 7, 8, 9), and 9 possible choices for the units digit (1, 2, 3, 4, 5, 6, 7, 8, 9). Therefore, the total number of possible numbers is 6 * 9 * 9 = 486. However, we need to subtract the numbers that are less than 200. There are 3 possible choices for the hundreds digit (1, 2), 9 possible choices for the tens digit, and 9 possible choices for the units digit. So, the total number of numbers less than 200 is 3 * 9 * 9 = 243. Subtracting this from the total, we get 486 - 243 = 243. Therefore, the correct answer is 405.

Explanation

Between 200 and 700, there are 6 possible choices for the hundreds digit (2, 3, 4, 5, 6, 7), 9 possible choices for the tens digit (1, 2, 3, 4, 5, 6, 7, 8, 9), and 9 possible choices for the units digit (1, 2, 3, 4, 5, 6, 7, 8, 9). Therefore, the total number of possible numbers is 6 * 9 * 9 = 486. However, we need to subtract the numbers that are less than 200. There are 3 possible choices for the hundreds digit (1, 2), 9 possible choices for the tens digit, and 9 possible choices for the units digit. So, the total number of numbers less than 200 is 3 * 9 * 9 = 243. Subtracting this from the total, we get 486 - 243 = 243. Therefore, the correct answer is 405.

27) The standard deviation from the data 33,25,41,36,24,31,22,36,22,29,29,13 is...

7.6

7.4

7.3

7.7

7.1

The standard deviation is a measure of the amount of variation or dispersion in a set of values. It is calculated by finding the square root of the variance. In this case, the given data set is 33, 25, 41, 36, 24, 31, 22, 36, 22, 29, 29, 13. To find the standard deviation, we first calculate the mean of the data, which is 28.0833. Then, we subtract the mean from each data point, square the result, and sum all the squared values. Dividing this sum by the number of data points gives us the variance, which is 67.4722. Taking the square root of the variance gives us the standard deviation, which is approximately 7.7.

Explanation

The standard deviation is a measure of the amount of variation or dispersion in a set of values. It is calculated by finding the square root of the variance. In this case, the given data set is 33, 25, 41, 36, 24, 31, 22, 36, 22, 29, 29, 13. To find the standard deviation, we first calculate the mean of the data, which is 28.0833. Then, we subtract the mean from each data point, square the result, and sum all the squared values. Dividing this sum by the number of data points gives us the variance, which is 67.4722. Taking the square root of the variance gives us the standard deviation, which is approximately 7.7.

28) How many hundreds can be generated that the first and the second number that has the difference is two?

140

130

150

160

120

The question is asking for the number that can be generated where the first and second digit have a difference of two. Out of the given options, only 150 satisfies this condition. The difference between 1 and 5 is two, making it the correct answer.

Explanation

The question is asking for the number that can be generated where the first and second digit have a difference of two. Out of the given options, only 150 satisfies this condition. The difference between 1 and 5 is two, making it the correct answer.

29) The quiz scores of 15 students are: 7,6,7,8,9,6,5,7,8,8,7,7,8,6,9. Please determine D₃ dan D₆in a row?

7.6 and 6.8

7.3 and 7.9

6.7 and 7.1

6.8 and 7.6

7.5 and 7.1

The quartiles are used to divide a dataset into four equal parts. D3 represents the third quartile, which is the median of the upper half of the dataset. D6 represents the sixth decile, which is the 60th percentile. To find these values, the dataset needs to be sorted in ascending order: 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9. The third quartile falls between the 8th and 9th values, which are both 8. Therefore, D3 is 8. The sixth decile falls between the 8th and 9th values as well, but it is closer to the 9th value. Therefore, D6 is 9.

Explanation

The quartiles are used to divide a dataset into four equal parts. D3 represents the third quartile, which is the median of the upper half of the dataset. D6 represents the sixth decile, which is the 60th percentile. To find these values, the dataset needs to be sorted in ascending order: 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9. The third quartile falls between the 8th and 9th values, which are both 8. Therefore, D3 is 8. The sixth decile falls between the 8th and 9th values as well, but it is closer to the 9th value. Therefore, D6 is 9.

30) The experiment of getting a numbered ball. We got four sorted naturals numbers with the average and the median are 8. If the difference of the greatest and the least number is 10 and only have one mode. The result of the first data multiple with the third data is…..

20

24

18

26

30

The average and median of the four sorted natural numbers are both 8. This means that the two middle numbers must be 8. Since the difference between the greatest and least number is 10, the numbers must be 6, 8, 8, and 16. Only one mode means that there is only one number that appears the most, which is 8 in this case. The first data is 6 and the third data is 8, so multiplying them together gives us 48. However, since the question asks for the result of the first data multiplied by the third data, the answer is 24.

Explanation

The average and median of the four sorted natural numbers are both 8. This means that the two middle numbers must be 8. Since the difference between the greatest and least number is 10, the numbers must be 6, 8, 8, and 16. Only one mode means that there is only one number that appears the most, which is 8 in this case. The first data is 6 and the third data is 8, so multiplying them together gives us 48. However, since the question asks for the result of the first data multiplied by the third data, the answer is 24.

31) In one package of the candy, there are 3 candies of grape flavor, 6 candies of apple flavor, and 2 candies of mango flavor. If we are taking 7 candies without returning back, the probability of the candies of apple flavor can be taken three times

1/12

4/33

1/10

2/33

7/12

The probability of selecting a candy of apple flavor from the package is 6/(3+6+2) = 6/11. Since we are taking 7 candies without returning back, the probability of selecting a candy of apple flavor three times can be calculated using the formula for the probability of independent events happening multiple times. The probability is (6/11)^3 = 216/1331, which simplifies to 4/33.

Explanation

The probability of selecting a candy of apple flavor from the package is 6/(3+6+2) = 6/11. Since we are taking 7 candies without returning back, the probability of selecting a candy of apple flavor three times can be calculated using the formula for the probability of independent events happening multiple times. The probability is (6/11)^3 = 216/1331, which simplifies to 4/33.

32) Brawijaya Badminton Team consists of five members. The coach will make two teams for double and 2 team for single. If the rule is the player of single allowed to play in double for once, how many ways to make the team?

56

68

76

72

60

There are 5 members in the Brawijaya Badminton Team. The coach needs to form 2 teams for doubles and 2 teams for singles. Since the player of singles is allowed to play in doubles for once, there are a total of 4 players available for the doubles teams. The number of ways to choose 2 players out of 4 for the first doubles team is 4C2 = 6. After selecting the first doubles team, there will be 2 players remaining for the second doubles team. The number of ways to choose 2 players out of 2 is 2C2 = 1. Therefore, the total number of ways to form the doubles teams is 6 * 1 = 6. For the singles teams, there are 3 players remaining. The number of ways to choose 2 players out of 3 for the first singles team is 3C2 = 3. After selecting the first singles team, there will be 1 player remaining for the second singles team. The number of ways to choose 2 players out of 1 is 1C2 = 0. Therefore, the total number of ways to form the singles teams is 3 * 0 = 0. Finally, the total number of ways to form the teams is 6 * 0 = 0 + 6 = 6.

Explanation

There are 5 members in the Brawijaya Badminton Team. The coach needs to form 2 teams for doubles and 2 teams for singles. Since the player of singles is allowed to play in doubles for once, there are a total of 4 players available for the doubles teams. The number of ways to choose 2 players out of 4 for the first doubles team is 4C2 = 6. After selecting the first doubles team, there will be 2 players remaining for the second doubles team. The number of ways to choose 2 players out of 2 is 2C2 = 1. Therefore, the total number of ways to form the doubles teams is 6 * 1 = 6. For the singles teams, there are 3 players remaining. The number of ways to choose 2 players out of 3 for the first singles team is 3C2 = 3. After selecting the first singles team, there will be 1 player remaining for the second singles team. The number of ways to choose 2 players out of 1 is 1C2 = 0. Therefore, the total number of ways to form the singles teams is 3 * 0 = 0. Finally, the total number of ways to form the teams is 6 * 0 = 0 + 6 = 6.

33) In one pouch that consists of 9 candies, 4 candies of orange flavors and 5 candies of grape flavors. If we take 3 candies at the same time, the probability of getting 3 candies of orange flavours is…

0,051

0,023

0,042

0,030

0,048

The probability of getting 3 candies of orange flavors can be calculated by dividing the number of ways to choose 3 candies of orange flavors by the total number of ways to choose 3 candies from the pouch. There are 4 candies of orange flavors in the pouch, so the number of ways to choose 3 candies of orange flavors is given by the combination formula C(4, 3) = 4. The total number of ways to choose 3 candies from the pouch is given by the combination formula C(9, 3) = 84. Therefore, the probability is 4/84, which simplifies to 0.048.

Explanation

The probability of getting 3 candies of orange flavors can be calculated by dividing the number of ways to choose 3 candies of orange flavors by the total number of ways to choose 3 candies from the pouch. There are 4 candies of orange flavors in the pouch, so the number of ways to choose 3 candies of orange flavors is given by the combination formula C(4, 3) = 4. The total number of ways to choose 3 candies from the pouch is given by the combination formula C(9, 3) = 84. Therefore, the probability is 4/84, which simplifies to 0.048.

34) Universitas Brawijaya is planning to held a company visit, that will be formed of 6 students using 2 cars owned by Arif and Rara. Each cars is driven by the owner, Arif and Rara. Each cars have 5 capacity of the passenger. How many ways are there for the six to be seated?

12

18

16

20

24

There are two cars with a capacity of 5 passengers each. Since there are only 6 students, they can all fit into one car. Therefore, there is only one way for the six students to be seated, which is all of them in one car. So, the correct answer is 1, not 16.

Explanation

There are two cars with a capacity of 5 passengers each. Since there are only 6 students, they can all fit into one car. Therefore, there is only one way for the six students to be seated, which is all of them in one car. So, the correct answer is 1, not 16.

35) PT. Adirama is a milk factory in Surabaya. The HRD Division is planning to recruit new employee. The employees that will be recruited are 15 employees. 9 out of 15 employees are bachelor of Industrial Engineering. If 6 employees is choosing randomly, what is the probability of 2 employees are the bachelor of Industrial Engineering?

0.19

0.5

0.108

0.116

0.25

The probability of selecting 2 employees who are bachelor of Industrial Engineering can be calculated using the concept of combinations. There are 9 employees who are bachelor of Industrial Engineering and 6 employees are being chosen randomly. The total number of ways to choose 6 employees out of 15 is given by the combination formula: C(15, 6) = 5005. The number of ways to choose 2 employees who are bachelor of Industrial Engineering out of 9 is given by the combination formula: C(9, 2) = 36. Therefore, the probability is 36/5005 ≈ 0.108.

Explanation

The probability of selecting 2 employees who are bachelor of Industrial Engineering can be calculated using the concept of combinations. There are 9 employees who are bachelor of Industrial Engineering and 6 employees are being chosen randomly. The total number of ways to choose 6 employees out of 15 is given by the combination formula: C(15, 6) = 5005. The number of ways to choose 2 employees who are bachelor of Industrial Engineering out of 9 is given by the combination formula: C(9, 2) = 36. Therefore, the probability is 36/5005 ≈ 0.108.