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Statistics and Probability - Measures of central tendency (mean, median, mode)

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Arithmetic Mean is the average of all data points, representing the sum of values divided by the number of values. Visually, the mean acts as the balance point or fulcrum of a data distribution; if you placed the data points on a see-saw, the mean is where the center support would be to keep it level.

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The Median is the central value of a data set when the values are arranged in order of size. Visually, if you imagine a line of students ordered by height, the median is the height of the student standing exactly in the middle. If there is an even number of data points, the median is the mid-point between the two central values.

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The Mode is the value that occurs most frequently in a distribution. In a visual representation like a bar chart or histogram, the mode is the value corresponding to the tallest bar. Data sets can be unimodal (one mode), bimodal (two modes), or have no mode if all values appear with the same frequency.

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For Grouped Data, the exact values are unknown, so we use the mid-interval value (xx) to estimate the mean. The Modal Class is the interval that contains the highest frequency. Visually, on a histogram, this is the widest or highest rectangular block representing a range of values.

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The Effect of Outliers refers to how extreme values impact measures of central tendency. The mean is 'sensitive' to outliers, meaning a single very high value will pull the mean upward. Visually, an outlier appears as a lonely point far to the left or right on a box plot or scatter diagram. The median is 'robust' because it is less affected by these extreme values.

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Cumulative Frequency Curves (Ogives) are used to visually estimate the median for grouped data. By plotting the cumulative frequency on the y-axis and the data values on the x-axis, the median is found by identifying the 50%50\% mark on the y-axis, moving horizontally to the curve, and then vertically down to the x-axis.

📐Formulae

xˉ=∑xn\bar{x} = \frac{\sum x}{n}

xˉ=∑fx∑f\bar{x} = \frac{\sum fx}{\sum f}

Median position=n+12\text{Median position} = \frac{n+1}{2}

Mid-interval value=lower bound+upper bound2\text{Mid-interval value} = \frac{\text{lower bound} + \text{upper bound}}{2}

💡Examples

Problem 1:

A student takes 5 math tests and scores: 72,85,78,90,8572, 85, 78, 90, 85. Find the mean, median, and mode of these scores.

Solution:

  1. Mean: xˉ=72+85+78+90+855=4105=82\bar{x} = \frac{72 + 85 + 78 + 90 + 85}{5} = \frac{410}{5} = 82. \ 2. Median: Arrange in order: 72,78,85,85,9072, 78, 85, 85, 90. The middle value (n=5n=5, position is 5+12=3\frac{5+1}{2} = 3rd) is 8585. \ 3. Mode: The value 8585 appears twice, while others appear once. Mode = 8585.

Explanation:

To find the mean, sum all scores and divide by the count. For the median, sorting is essential to find the middle position. The mode is simply the most frequent score.

Problem 2:

Estimate the mean for the following grouped frequency table: \ Intervals: 0≤x<100 \le x < 10 (Freq: 33), 10≤x<2010 \le x < 20 (Freq: 55), 20≤x<3020 \le x < 30 (Freq: 22)

Solution:

  1. Find midpoints (xx): 5,15,255, 15, 25. \ 2. Multiply midpoints by frequencies (fxfx): (5×3)=15(5 \times 3) = 15; (15×5)=75(15 \times 5) = 75; (25×2)=50(25 \times 2) = 50. \ 3. Sum of fxfx: ∑fx=15+75+50=140\sum fx = 15 + 75 + 50 = 140. \ 4. Sum of ff: ∑f=3+5+2=10\sum f = 3 + 5 + 2 = 10. \ 5. Estimate Mean: xˉ=14010=14\bar{x} = \frac{140}{10} = 14.

Explanation:

Because we don't know the exact values within the intervals, we use the midpoint of each class as a representative value. The estimated mean is the total of these products divided by the total frequency.