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Statistics and Probability - Measures of central tendency

Grade 11IB_AI

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

🔑Concepts

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The Mean (xˉ\bar{x}) is the arithmetic average of a data set. It is sensitive to outliers and extreme values.

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The Median (MM or Q2Q_2) is the middle value when the data is arranged in ascending or descending order. It is a robust measure as it is less affected by outliers.

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The Mode is the most frequently occurring value in a data set. A data set can be bimodal (two modes) or have no mode at all.

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For Grouped Data, we estimate the mean by using the midpoint (xx) of each class interval and multiplying it by the frequency (ff).

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In a Symmetrical Distribution, the mean, median, and mode are approximately equal. In a Skewed Distribution, the mean is pulled toward the tail.

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For the IB AI course, the Graphic Display Calculator (GDC) is the primary tool for calculating these measures using '1-Variable Statistics' functions.

📐Formulae

xˉ=∑i=1nxin\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}

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

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

💡Examples

Problem 1:

Calculate the mean and median for the following discrete frequency table: Value (x)5101520Frequency (f)2521\begin{array}{|c|c|c|c|c|} \hline \text{Value } (x) & 5 & 10 & 15 & 20 \\ \hline \text{Frequency } (f) & 2 & 5 & 2 & 1 \\ \hline \end{array}

Solution:

  1. Find ∑f\sum f: 2+5+2+1=102 + 5 + 2 + 1 = 10.
  2. Find ∑fx\sum fx: (5×2)+(10×5)+(15×2)+(20×1)=10+50+30+20=110(5 \times 2) + (10 \times 5) + (15 \times 2) + (20 \times 1) = 10 + 50 + 30 + 20 = 110.
  3. Mean xˉ=11010=11\bar{x} = \frac{110}{10} = 11.
  4. Median Position: 10+12=5.5th\frac{10+1}{2} = 5.5^{th} value.
  5. The cumulative frequencies are 2,7,9,102, 7, 9, 10. The 5.5th5.5^{th} value falls in the second group, so Median = 1010.

Explanation:

We use the weighted mean formula for frequency tables. The median is found by identifying the value corresponding to the middle position in the cumulative frequency.

Problem 2:

In a group of 5 students, the heights are 160160 cm, 165165 cm, 170170 cm, 175175 cm, and xx cm. If the mean height is 168168 cm, find the value of xx.

Solution:

Using the mean formula: 160+165+170+175+x5=168\frac{160 + 165 + 170 + 175 + x}{5} = 168 670+x=168×5670 + x = 168 \times 5 670+x=840670 + x = 840 x=840−670x = 840 - 670 x=170x = 170

Explanation:

To find a missing value, we set up an algebraic equation based on the definition of the arithmetic mean and solve for the unknown xx.

Problem 3:

Consider the following grouped frequency table for the weights (ww) of 20 items: Weight (w, kg)Frequency (f)0≤w<10410≤w<201220≤w<304\begin{array}{|c|c|} \hline \text{Weight } (w, \text{ kg}) & \text{Frequency } (f) \\ \hline 0 \le w < 10 & 4 \\ 10 \le w < 20 & 12 \\ 20 \le w < 30 & 4 \\ \hline \end{array} Estimate the mean weight.

Solution:

  1. Identify midpoints (xx): 5,15,255, 15, 25.
  2. Calculate fxfx:
    • 5×4=205 \times 4 = 20
    • 15×12=18015 \times 12 = 180
    • 25×4=10025 \times 4 = 100
  3. ∑fx=20+180+100=300\sum fx = 20 + 180 + 100 = 300.
  4. ∑f=4+12+4=20\sum f = 4 + 12 + 4 = 20.
  5. Estimated Mean xˉ=30020=15\bar{x} = \frac{300}{20} = 15 kg.

Explanation:

For grouped data, we assume all values in an interval are located at the midpoint. This provides an estimate of the mean.