Math Problem Statement

Calculate the Upper Quartile (𝑄3) and the Lower Quartile (𝑄1) for the following frequency distribution. And hence calculate the quartile deviation and the coefficient of quartile deviation. Class Interval: 0 – 8 8 – 16 16 – 24 8 – 32 32 – 40 Frequency: 20 12 25 15 8

Solution

Step 1: Determine the Cumulative Frequency

The first step is to compute the cumulative frequency for the given data.

Class IntervalFrequency (f)Cumulative Frequency (cf)
0 – 82020
8 – 161232
16 – 242557
24 – 321572
32 – 40880

Step 2: Identify Quartiles

The quartiles divide the dataset into four equal parts. Since there are 80 total frequencies, we calculate the positions for Q1Q1 and Q3Q3 as follows:

  • Q1Q1 (Lower Quartile) is at position 14Γ—80=20\frac{1}{4} \times 80 = 20
  • Q3Q3 (Upper Quartile) is at position 34Γ—80=60\frac{3}{4} \times 80 = 60

Step 3: Calculate Q1Q1 (Lower Quartile)

Q1Q1 is at the 20th position, which falls in the class interval 0 – 8 (since the cumulative frequency up to this class is 20). We use the quartile formula:

Q1=L+(N4βˆ’cfprevf)Γ—hQ1 = L + \left( \frac{\frac{N}{4} - cf_{\text{prev}}}{f} \right) \times h

Where:

  • LL = lower boundary of the class containing Q1Q1 = 0
  • cfprevcf_{\text{prev}} = cumulative frequency before the class of Q1Q1 = 0
  • ff = frequency of the class containing Q1Q1 = 20
  • N=80N = 80
  • hh = class width = 8

Substitute the values:

Q1=0+(20βˆ’020)Γ—8=0+8=8Q1 = 0 + \left( \frac{20 - 0}{20} \right) \times 8 = 0 + 8 = 8

So, Q1=8Q1 = 8.

Step 4: Calculate Q3Q3 (Upper Quartile)

Q3Q3 is at the 60th position, which falls in the class interval 24 – 32 (since the cumulative frequency up to the previous class is 57 and for this class it reaches 72). We use the same quartile formula:

Q3=L+(3N4βˆ’cfprevf)Γ—hQ3 = L + \left( \frac{\frac{3N}{4} - cf_{\text{prev}}}{f} \right) \times h

Where:

  • LL = lower boundary of the class containing Q3Q3 = 24
  • cfprevcf_{\text{prev}} = cumulative frequency before the class of Q3Q3 = 57
  • ff = frequency of the class containing Q3Q3 = 15
  • hh = class width = 8

Substitute the values:

Q3=24+(60βˆ’5715)Γ—8=24+(315)Γ—8=24+1.6=25.6Q3 = 24 + \left( \frac{60 - 57}{15} \right) \times 8 = 24 + \left( \frac{3}{15} \right) \times 8 = 24 + 1.6 = 25.6

So, Q3=25.6Q3 = 25.6.

Step 5: Calculate Quartile Deviation

The Quartile Deviation (QD) is calculated as:

QD=Q3βˆ’Q12=25.6βˆ’82=17.62=8.8QD = \frac{Q3 - Q1}{2} = \frac{25.6 - 8}{2} = \frac{17.6}{2} = 8.8

Step 6: Calculate Coefficient of Quartile Deviation

The Coefficient of Quartile Deviation (CQD) is calculated as:

CQD=Q3βˆ’Q1Q3+Q1=25.6βˆ’825.6+8=17.633.6β‰ˆ0.524CQD = \frac{Q3 - Q1}{Q3 + Q1} = \frac{25.6 - 8}{25.6 + 8} = \frac{17.6}{33.6} \approx 0.524

Final Results

  • Q1=8Q1 = 8
  • Q3=25.6Q3 = 25.6
  • Quartile Deviation =8.8= 8.8
  • Coefficient of Quartile Deviation β‰ˆ0.524\approx 0.524

Do you want more details on any step or have any questions?

Here are 5 related questions to explore further:

  1. How is the quartile deviation different from the standard deviation?
  2. How would you calculate the median for this frequency distribution?
  3. What would the effect of a smaller class width be on quartiles?
  4. How is the interquartile range (IQR) related to quartile deviation?
  5. How can we interpret the coefficient of quartile deviation in real-world data?

Tip: Quartiles are often used to measure the spread of data, making them helpful in understanding variability and identifying outliers.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Quartiles
Cumulative Frequency
Quartile Deviation
Coefficient of Quartile Deviation

Formulas

Q1 = L + [(N/4 - cf_prev) / f] * h
Q3 = L + [(3N/4 - cf_prev) / f] * h
Quartile Deviation = (Q3 - Q1) / 2
Coefficient of Quartile Deviation = (Q3 - Q1) / (Q3 + Q1)

Theorems

Quartile Calculation
Cumulative Frequency Theorem

Suitable Grade Level

Grades 10-12