krit.club logo

Statistics and Probability - Discrete and Continuous Data, Grouped and Ungrouped

Grade 8IB

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

🔑Concepts

•

Discrete data can only take specific, distinct values, often as a result of counting (e.g., number of students n=25n = 25).

•

Continuous data can take any value within a given range and is usually the result of measuring (e.g., height hh such that 160≤h<170160 \le h < 170 cm).

•

Ungrouped data is listed as individual values or in a frequency table where each value is treated separately.

•

Grouped data organizes large datasets into class intervals (e.g., 10−20,20−3010 - 20, 20 - 30).

•

For grouped data, the 'Midpoint' (mm) of a class interval is used to estimate the mean.

•

The Modal Class is the interval with the highest frequency in a grouped frequency table.

•

The Median of nn ordered data points is found at the n+12\frac{n+1}{2} position.

📐Formulae

Mean for ungrouped data (xˉ)=∑xn\text{Mean for ungrouped data } (\bar{x}) = \frac{\sum x}{n}

Mean from a frequency table (xˉ)=∑(f⋅x)∑f\text{Mean from a frequency table } (\bar{x}) = \frac{\sum (f \cdot x)}{\sum f}

Midpoint (m)=Lower Limit+Upper Limit2\text{Midpoint } (m) = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}

Estimated Mean (Grouped)=∑(f⋅m)∑f\text{Estimated Mean (Grouped)} = \frac{\sum (f \cdot m)}{\sum f}

Range=Maximum Value−Minimum Value\text{Range} = \text{Maximum Value} - \text{Minimum Value}

💡Examples

Problem 1:

Find the mean of the following discrete data representing the number of goals scored in 5 matches: 2,0,3,3,22, 0, 3, 3, 2.

Solution:

xˉ=2+0+3+3+25=105=2\bar{x} = \frac{2 + 0 + 3 + 3 + 2}{5} = \frac{10}{5} = 2

Explanation:

To find the mean of ungrouped data, add all the values and divide by the total number of observations nn.

Problem 2:

Given the following frequency table for the number of pets (xx) owned by students (ff): x=0,f=2x=0, f=2; x=1,f=5x=1, f=5; x=2,f=3x=2, f=3. Calculate the mean.

Solution:

∑f=2+5+3=10\sum f = 2 + 5 + 3 = 10 ∑(f⋅x)=(2×0)+(5×1)+(3×2)=0+5+6=11\sum (f \cdot x) = (2 \times 0) + (5 \times 1) + (3 \times 2) = 0 + 5 + 6 = 11 xˉ=1110=1.1\bar{x} = \frac{11}{10} = 1.1

Explanation:

Multiply each value xx by its frequency ff, sum them up, and divide by the total frequency.

Problem 3:

Estimate the mean weight of apples from the following grouped data: 100≤w<120100 \le w < 120 g: frequency 44 120≤w<140120 \le w < 140 g: frequency 66

Solution:

Midpoint m1=100+1202=110m_1 = \frac{100+120}{2} = 110 Midpoint m2=120+1402=130m_2 = \frac{120+140}{2} = 130 Estimated Mean=(4×110)+(6×130)4+6\text{Estimated Mean} = \frac{(4 \times 110) + (6 \times 130)}{4 + 6} Estimated Mean=440+78010=122010=122 g\text{Estimated Mean} = \frac{440 + 780}{10} = \frac{1220}{10} = 122 \text{ g}

Explanation:

For continuous grouped data, find the midpoint of each class, multiply by the frequency, and divide the total sum by the sum of frequencies.