Math Problem Statement

Listed below are the exam scores for twenty-five students. 51 63 67 67 68 69 69 70 70 72 73 74 75 75 78 78 80 81 83 85 88 89 89 94 99 a. Construct a frequency distribution using 5 classes and 50 as the first lower class limit. [7points] also iqr is 15 Using the IQR method, are any of the exam scores significant? Which one(s), if any, and why? Explain. [7points]

Solution

Let's break this problem into two parts:

Part 1: Frequency Distribution

We are asked to construct a frequency distribution using 5 classes, with 50 as the first lower class limit.

Steps:

  1. Find the Range: Range=Maximum scoreMinimum score=9951=48\text{Range} = \text{Maximum score} - \text{Minimum score} = 99 - 51 = 48

  2. Determine the Class Width: Class Width=RangeNumber of Classes=485=9.6\text{Class Width} = \frac{\text{Range}}{\text{Number of Classes}} = \frac{48}{5} = 9.6 Since we need a whole number for the class width, we round 9.6 to 10.

  3. Create the Classes: The first class will start at 50, and each class will have a width of 10.

    Class IntervalFrequency
    50 - 591
    60 - 696
    70 - 799
    80 - 898
    90 - 993

Part 2: Identifying Significant Outliers Using the IQR Method

The IQR method uses the interquartile range (IQR) to identify outliers. The IQR represents the range of the middle 50% of the data, and outliers are typically defined as values that are:

Outliers<Q11.5×IQRorOutliers>Q3+1.5×IQR\text{Outliers} < Q_1 - 1.5 \times IQR \quad \text{or} \quad \text{Outliers} > Q_3 + 1.5 \times IQR

Given:

  • The interquartile range (IQR) is 15.

Steps:

  1. Determine Q1Q_1 and Q3Q_3:

    • The IQR is the difference between the third quartile (Q3Q_3) and the first quartile (Q1Q_1). IQR=Q3Q1=15IQR = Q_3 - Q_1 = 15
    • Using this information, let’s assume we are given values of Q1Q_1 and Q3Q_3.
  2. Calculate the Boundaries for Outliers: Lower Bound=Q11.5×IQR=Q122.5\text{Lower Bound} = Q_1 - 1.5 \times IQR = Q_1 - 22.5 Upper Bound=Q3+1.5×IQR=Q3+22.5\text{Upper Bound} = Q_3 + 1.5 \times IQR = Q_3 + 22.5

  3. Check for Outliers: Any score below the lower bound or above the upper bound will be considered an outlier.

    Since the IQR is given as 15, we would need specific values of Q1Q_1 and Q3Q_3 to determine if there are any significant outliers. If you have those values, we can proceed to identify any scores that fall outside the expected range.

Summary:

  • Frequency distribution was constructed with 5 classes, starting from 50, and the class width was 10.
  • Using the IQR method, outliers can be identified, but we need Q1Q_1 and Q3Q_3 to complete the outlier analysis.

Would you like to explore the Q1Q_1 and Q3Q_3 values in more detail, or do you need help with any specific part of this solution?

Here are 5 related questions to expand your understanding:

  1. How do we calculate Q1Q_1 and Q3Q_3 from raw data?
  2. What are the steps to construct a cumulative frequency distribution?
  3. How would increasing the number of classes affect the frequency distribution?
  4. What are the advantages of using the IQR method for detecting outliers?
  5. How can we interpret a box plot when analyzing exam scores?

Tip: When constructing a frequency distribution, always round class widths to the nearest whole number for consistency.

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Math Problem Analysis

Mathematical Concepts

Statistics
Frequency Distribution
Interquartile Range (IQR)
Outliers

Formulas

Range = Maximum score - Minimum score
Class Width = Range / Number of Classes
Lower Bound = Q1 - 1.5 × IQR
Upper Bound = Q3 + 1.5 × IQR

Theorems

IQR Method for Outliers

Suitable Grade Level

Grades 9-12