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Statistics and Probability - Basic concepts of Statistics

Grade 11IB_AA

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

🔑Concepts

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Types of Data: Quantitative data can be discrete (countable, like the number of students) or continuous (measurable, like height or time). Qualitative data is categorical (like eye color).

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Sampling Techniques: Methods used to select a subset of a population. Common methods include Simple Random Sampling (each member has an equal chance), Systematic Sampling (choosing every kthk^{th} member), and Stratified Sampling (choosing proportional samples from subgroups).

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Measures of Central Tendency: The Mean xˉ\bar{x} is the arithmetic average. The Median is the middle value when data is ordered. The Mode is the most frequent value.

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Measures of Dispersion: These describe the spread of data. The Range is the difference between the maximum and minimum values. The Interquartile Range (IQRIQR) is the spread of the middle 50%50\% of data.

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Variance and Standard Deviation: The standard deviation σ\sigma measures the average distance of data points from the mean. A low σ\sigma indicates data is close to the mean.

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Outliers: Data points that fall significantly outside the rest of the dataset. They are mathematically defined using the 1.5×IQR1.5 \times IQR rule.

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Cumulative Frequency: The running total of frequencies. Cumulative frequency graphs (ogives) are used to estimate the median, quartiles, and percentiles.

📐Formulae

xˉ=∑i=1kfixin\bar{x} = \frac{\sum_{i=1}^{k} f_i x_i}{n}

σ2=∑i=1kfi(xi−xˉ)2n\sigma^2 = \frac{\sum_{i=1}^{k} f_i (x_i - \bar{x})^2}{n}

σ=∑i=1kfi(xi−xˉ)2n\sigma = \sqrt{\frac{\sum_{i=1}^{k} f_i (x_i - \bar{x})^2}{n}}

IQR=Q3−Q1IQR = Q_3 - Q_1

Lower Outlier Boundary=Q1−1.5×IQR\text{Lower Outlier Boundary} = Q_1 - 1.5 \times IQR

Upper Outlier Boundary=Q3+1.5×IQR\text{Upper Outlier Boundary} = Q_3 + 1.5 \times IQR

💡Examples

Problem 1:

Given the data set: 3,7,8,5,12,14,21,15,183, 7, 8, 5, 12, 14, 21, 15, 18, find the median and the Interquartile Range (IQRIQR).

Solution:

  1. Order the data: 3,5,7,8,12,14,15,18,213, 5, 7, 8, 12, 14, 15, 18, 21.
  2. The number of values n=9n = 9.
  3. Median (Q2Q_2): The 9+12=5th\frac{9+1}{2} = 5^{th} value, which is 1212.
  4. Lower Quartile (Q1Q_1): The median of the lower half (3,5,7,83, 5, 7, 8) is 5+72=6\frac{5+7}{2} = 6.
  5. Upper Quartile (Q3Q_3): The median of the upper half (14,15,18,2114, 15, 18, 21) is 15+182=16.5\frac{15+18}{2} = 16.5.
  6. IQR=Q3−Q1=16.5−6=10.5IQR = Q_3 - Q_1 = 16.5 - 6 = 10.5.

Explanation:

To find quartiles, first arrange data in ascending order. If nn is odd, the median is the middle term. Q1Q_1 and Q3Q_3 are the medians of the two halves created by the median.

Problem 2:

A data set has Q1=10Q_1 = 10 and Q3=25Q_3 = 25. Determine if the value 5050 is an outlier.

Solution:

  1. Calculate IQRIQR: IQR=Q3−Q1=25−10=15IQR = Q_3 - Q_1 = 25 - 10 = 15.
  2. Calculate the upper outlier boundary: Q3+1.5×IQR=25+1.5(15)Q_3 + 1.5 \times IQR = 25 + 1.5(15).
  3. 25+22.5=47.525 + 22.5 = 47.5.
  4. Compare the value: Since 50>47.550 > 47.5, the value 5050 is an outlier.

Explanation:

An outlier is any value that is greater than the upper boundary (Q3+1.5×IQRQ_3 + 1.5 \times IQR) or smaller than the lower boundary (Q1−1.5×IQRQ_1 - 1.5 \times IQR).

Problem 3:

Calculate the mean of the following frequency distribution:

Value (x)Frequency (f)102203305\begin{array}{|c|c|} \hline \text{Value } (x) & \text{Frequency } (f) \\ \hline 10 & 2 \\ 20 & 3 \\ 30 & 5 \\ \hline \end{array}

Solution:

  1. Calculate ∑fixi\sum f_i x_i: (10×2)+(20×3)+(30×5)=20+60+150=230(10 \times 2) + (20 \times 3) + (30 \times 5) = 20 + 60 + 150 = 230.
  2. Calculate nn (total frequency): 2+3+5=102 + 3 + 5 = 10.
  3. Mean xˉ=23010=23\bar{x} = \frac{230}{10} = 23.

Explanation:

For frequency distributions, the mean is the sum of the products of each value and its frequency, divided by the total frequency.