Statistics (Central Tendency) MCQ Test — Class 7 IB MYP

Solution:

Sum all the given temperatures. — Sum = 18 + 22 + 19 + 21 + 25 = 105

Count the number of days (data values). — Number of values = 5

Divide the sum by the number of values to find the mean. — Mean = Sum / Number of values = 105 / 5 = 21

Solution:

Recall the definition of the median: the middle value in an ordered data set.

Check Alex's answer against the original, unordered data. If 18 is the median, it must be the middle value after ordering. — Original: 15, 12, 20, 18, 14, 16

Order the numbers: 12, 14, 15, 16, 18, 20. The middle two numbers are 15 and 16. The median is (15 + 16) / 2 = 15.5.

Compare Alex's answer (18) with the correct median (15.5) and the required steps. Alex's mistake was not ordering the numbers before selecting a middle value.

Solution:

Recall the definition of the mode: the value that occurs most often in a data set.

Evaluate option A: A data set can have no mode (if all values appear once) or multiple modes (if two or more values share the highest frequency). So, A is false.

Evaluate option B: The mode can be any value in the data set, not necessarily the largest. So, B is false.

Evaluate option C: This matches the definition of the mode. So, C is true.

Evaluate option D: The mode is generally not significantly affected by outliers, as it only depends on the frequency of values, not their magnitude relative to the rest of the data. So, D is false.

Solution:

Consider the mean: It is calculated by summing all values and dividing by the count. An extremely high value (outlier) will significantly increase the sum, thus heavily influencing the mean.

Consider the median: It is the middle value when data is ordered. While an outlier might shift the ordering slightly, its extreme value doesn't directly pull the median itself as much as the mean.

Consider the mode: It is the most frequent value. An outlier is a unique, extreme value, so it typically has no impact on the mode unless it happens to be the most frequent value, which is unlikely for an outlier.

Therefore, the mean is the measure most sensitive to extreme values or outliers.

Solution:

The mean provides the average score but doesn't highlight the most frequent score.

The median provides the middle score, which may not be the most common.

The mode specifically identifies the value that appears most often in a data set. This directly answers 'the most common score'.

The range measures the spread of data, not a central tendency.

Therefore, the mode is the most appropriate measure to show the most common score.

Solution:

Identify the mode: The 'Number of siblings' with the highest 'Frequency' is 1 (with 8 students). So, Mode = 1. — Frequency Table: Siblings (x) | Frequency (f) 0 | 3 1 | 8 2 | 5 3 | 2 4 | 1

Calculate the total number of students to find the median position. — Total students = 3 + 8 + 5 + 2 + 1 = 19

For 19 data points, the median is the (19 + 1) / 2 = 10th value.

List the data in order from the frequency table: 0, 0, 0 (3 times); 1, 1, ..., 1 (8 times); 2, 2, ..., 2 (5 times); etc. The 10th value falls within the '1 sibling' category (after the 3 zeros, the next 8 values are ones). — Cumulative frequency: 0 (3), 1 (3+8=11), 2 (11+5=16), 3 (16+2=18), 4 (18+1=19). The 10th value is 1.

So, the median is 1.

Solution:

List all the individual scores from the stem-and-leaf plot. — Scores = 21, 24, 25, 30, 32, 32, 37, 41, 45, 48, 49

Count the number of students (data points). There are 12 leaf values, so there are 12 scores. — Number of scores = 12

Sum all the scores. — Sum = 21 + 24 + 25 + 30 + 32 + 32 + 37 + 41 + 45 + 48 + 49 = 394

Calculate the mean by dividing the sum by the number of scores. — Mean = 394 / 12 = 32.83 (Wait, mistake in data copy). Let's re-list and re-sum carefully. Scores: 21, 24, 25, 30, 32, 32, 37, 41, 45, 48, 49. (11 scores) Ah, the question says 12 students, but only 11 leaves are listed. Let me add one more score to make it 12, or adjust the count. The question states 12 students. So, I will assume there's one missing leaf in the example provided, or re-count. Let's assume the question text implies 12 students and the listed stem-and-leaf plot has 12 values. Stem | Leaf 2 | 1 4 5 3 | 0 2 2 7 4 | 1 5 8 9 (This is 11 leaves, let's add one, say 35 to make it 12) Let's assume the original question meant 11 students with the given plot, or I should adjust the data to have 12. For consistency, let's use the given 11 values and adjust the problem statement's count if needed, or make sure my options match 11 or 12. If 11 scores: 21+24+25+30+32+32+37+41+45+48+49 = 384. Mean = 384/11 = 34.9 (approx). This is close to 34.5 or 35.0. Let's assume I missed one score when creating. I will re-create the data set to have 12 scores and a clean mean. New data: Stem | Leaf 2 | 1 4 5 3 | 0 2 2 5 7 4 | 1 5 8 9 Scores: 21, 24, 25, 30, 32, 32, 35, 37, 41, 45, 48, 49. (12 scores) Sum = 21+24+25+30+32+32+35+37+41+45+48+49 = 420 Mean = 420 / 12 = 35.0. This is one of the options. Okay, I will use this revised plot.

Solution:

Let 'x' be the score in the fifth game. The total points for five games will be the sum of the four known scores plus x. — Total points = 12 + 18 + 15 + 20 + x = 65 + x

The desired mean for five games is 17 points. Set up the equation for the mean. — (65 + x) / 5 = 17

Solve the equation for x. — 65 + x = 17 × 5

Calculate the value of x. — 65 + x = 85 x = 85 - 65 x = 20

Solution:

Evaluate option A: For grouped data, we can only estimate the mean, not find it exactly, as individual data values are unknown. So, A is incorrect.

Evaluate option B: The modal class is the class interval with the highest frequency. The 150-159 cm class has 7 students, which is the highest frequency. So, B is correct.

Evaluate option C: For grouped data, we can only identify the class that contains the median, not the exact median value itself. So, C is incorrect.

Evaluate option D: Students taller than 170 cm (i.e., in the 170-179 cm class) are 2. Students shorter than 150 cm (i.e., in the 140-149 cm class) are 4. So, 2 is not more than 4. D is incorrect.

Solution:

Consistency implies how spread out or varied the data is, or how well a typical value represents the set. The mode simply tells us the most frequent rainfall amount.

If many days have 0 mm rainfall, the mode might be 0, which doesn't reflect the overall amount or consistency of rain on rainy days.

The mean and median provide a central value that can give a sense of the 'average' rainfall, which is more useful for comparing overall amounts.

For consistency, measures of spread like range or interquartile range would be most useful, but among central tendency measures, the mode is often the least informative for assessing overall consistency.