International Statistical Literacy Competition Quiz!

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.