Measure of Location | Jamb Mathematics

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📢 Yo, listen up! 🎤 Your exam on Measures of Central Tendency is coming up, and you gotta be ready! 🔥 Master mean, median, mode, and know when to use each—because stats don’t lie, but bad prep does! 📊💡 Now go crush those numbers like a pro! 🚀✅ #StudyMode #MathWhiz

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Measure of Location? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Measure of location together and move one step closer to achieving your exam success! Blissful learning.

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Calculation problem involving mean mode and median of ungrouped data

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Problem 1: Finding the Mean

Find the mean of the numbers 5, 8, 12, 15, 10.

Solution:

Use the formula for the mean: $\text{Mean} = \frac{\sum x_i}{n}$
Compute: $\text{Mean} = \frac{5 + 8 + 12 + 15 + 10}{5} = \frac{50}{5} = 10$ .
Final Answer: 10.

Problem 2: Finding the Median (Odd Set)

Find the median of 3, 7, 1, 9, 5.

Solution:

Arrange in ascending order: 1, 3, 5, 7, 9.
Since there are 5 numbers (odd count), the median is the middle number (3rd position).
Final Answer: 5.

Problem 3: Finding the Median (Even Set)

Find the median of 6, 2, 9, 4, 8, 5.

Solution:

Arrange in ascending order: 2, 4, 5, 6, 8, 9.
Since there are 6 numbers (even count), the median is the average of the 3rd and 4th numbers. $\text{Median} = \frac{5 + 6}{2} = \frac{11}{2} = 5.5$ .
Final Answer: 5.5.

Problem 4: Finding the Mode

Find the mode of 4, 8, 6, 4, 9, 8, 4.

Solution:

Count occurrences:
- 4 appears 3 times,
- 8 appears 2 times,
- 6 and 9 appear once.
The most frequent number is 4.
Final Answer: 4.

Problem 5: Mean with Frequencies

Find the mean of the dataset:

| Value (

x_i

) | Frequency (

f_i

) | |--------------|--------------| | 2 | 3 | | 5 | 4 | | 7 | 2 | | 10 | 1 |

Solution:

Compute $\sum f_i x_i$ : $(2 \times 3) + (5 \times 4) + (7 \times 2) + (10 \times 1) = 6 + 20 + 14 + 10 = 50$ .
Compute total frequency:
$n = 3 + 4 + 2 + 1 = 10$ .
Compute mean:
$\text{Mean} = \frac{50}{10} = 5$ .
Final Answer: 5.

Problem 6: Median in a Large Data Set

Find the median of 3, 7, 12, 15, 21, 25, 30, 35, 40, 50.

Solution:

Data is already sorted.
Since there are 10 numbers (even count), median is: $\frac{21 + 25}{2} = \frac{46}{2} = 23$ .
Final Answer: 23.

Problem 7: Mode for Bimodal Data

Find the mode of 2, 5, 7, 2, 5, 8, 9, 5, 2.

Solution:

Count occurrences:
- 2 appears 3 times,
- 5 appears 3 times,
- 7, 8, and 9 appear once.
Since 2 and 5 both appear most frequently, the data is bimodal.
Final Answer: 2 and 5.

Problem 8: Mean of Negative Numbers

Find the mean of -4, -8, -2, -10, -6.

Solution:

Compute sum:
$(-4) + (-8) + (-2) + (-10) + (-6) = -30$ .
Compute mean:
$\text{Mean} = \frac{-30}{5} = -6$ .
Final Answer: -6.

Problem 9: Median in a Skewed Distribution

Find the median of 3, 5, 8, 15, 22, 30, 100.

Solution:

Data is sorted.
Since there are 7 numbers (odd count), the median is the middle value (4th position).
Final Answer: 15.

Problem 10: Mode in a Frequency Table

Find the mode from:

| Number | Frequency | |---------|----------| | 1 | 2 | | 2 | 4 | | 3 | 3 | | 4 | 4 | | 5 | 1 |

Solution:

The highest frequency is 4, which appears for 2 and 4.
Final Answer: 2 and 4 (bimodal).

Problem 11: Finding the Mean of a Decimals Dataset

Find the mean of 2.5, 3.2, 4.8, 5.6, 6.0.

Solution:

Compute sum:
$2.5 + 3.2 + 4.8 + 5.6 + 6.0 = 22.1$ .
Compute mean:
$\text{Mean} = \frac{22.1}{5} = 4.42$ .
Final Answer: 4.42.

Problem 12: Median of a Large Dataset

Find the median of 20, 25, 30, 35, 40, 45, 50, 55, 60.

Solution:

Data is already sorted.
Since there are 9 numbers (odd count), the median is the middle value (5th position).
Final Answer: 40.

Problem 13: Finding the Mode with No Repetition

Find the mode of 1, 3, 5, 7, 9.

Solution:

Each number appears once.
Since no number repeats, there is no mode.
Final Answer: No mode.

Problem 14: Mean of a Weighted Dataset

A teacher calculates grades with weights:

| Score | Weight | |--------|--------| | 80 | 3 | | 90 | 5 | | 85 | 2 |

Find the weighted mean.

Solution:

Compute weighted sum:
$(80 \times 3) + (90 \times 5) + (85 \times 2) = 240 + 450 + 170 = 860$ .
Compute total weight:
$3 + 5 + 2 = 10$ .
Compute weighted mean:
$\frac{860}{10} = 86$ .
Final Answer: 86.

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- Jamb Mathematics- Lesson notes on Measure of Dispersion for utme Success

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This is all we can take on "Jamb Mathematics - Lesson Notes on Measure of Location for UTME Candidate"

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Measure of Location | Jamb Mathematics

Calculation problem involving mean mode and median of ungrouped data

Problem 1: Finding the Mean

Problem 2: Finding the Median (Odd Set)

Problem 3: Finding the Median (Even Set)

Problem 4: Finding the Mode

Problem 5: Mean with Frequencies

Problem 6: Median in a Large Data Set

Problem 7: Mode for Bimodal Data

Problem 8: Mean of Negative Numbers

Problem 9: Median in a Skewed Distribution

Problem 10: Mode in a Frequency Table

Problem 11: Finding the Mean of a Decimals Dataset

Problem 12: Median of a Large Dataset

Problem 13: Finding the Mode with No Repetition

Problem 14: Mean of a Weighted Dataset

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POSCHOLARS TEAM.

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