Math Problem Statement

4 | x 7 → 47 minutes

5 | 0 1 2 4 4 6 7 9 → 50, 51, 52, 54, 54, 56, 57, 59 minutes

6 | 0 2 3 6 7 → 60, 62, 63, 66, 67 minutes

7 | 1 2 3 7 → 71, 72, 73, 77 minutes x is there,infront of the number '7' at the 40 row

Solution

Thank you for the clarification. Since xx is in front of the number '7' in the "40s" row, it means the possible value for xx is somewhere between 40 and 49 minutes.

Given this information, let's reanalyze the problem:

Step 1: Revised Data Interpretation

The stem-and-leaf diagram now shows:

  • 4 | x 74x4x (where xx is a number between 0 and 9) and 47 minutes.
  • 5 | 0 1 2 4 4 6 7 9 → 50, 51, 52, 54, 54, 56, 57, 59 minutes.
  • 6 | 0 2 3 6 7 → 60, 62, 63, 66, 67 minutes.
  • 7 | 1 2 3 7 → 71, 72, 73, 77 minutes.

Step 2: List the Data in Ascending Order

The values now are: 4x,47,50,51,52,54,54,56,57,59,60,62,63,66,67,71,72,73,774x, 47, 50, 51, 52, 54, 54, 56, 57, 59, 60, 62, 63, 66, 67, 71, 72, 73, 77

Step 3: Range Calculation

Since xx is between 0 and 9, the range is still: Range=774x\text{Range} = 77 - 4x

Step 4: Given Condition

The interquartile range (IQR) is given as half of the total range.

Step 5: Calculate Quartiles

To calculate the quartiles, let's first assume some possible values for xx and then calculate the IQR.

Case 1: x=0x = 0 (i.e., 4040 minutes)

  • Data: 40, 47, 50, 51, 52, 54, 54, 56, 57, 59, 60, 62, 63, 66, 67, 71, 72, 73, 77
  • Median (Q2) = Average of the 9th and 10th values: Median (Q2)=57+592=58 minutes\text{Median (Q2)} = \frac{57 + 59}{2} = 58 \text{ minutes}
  • Lower Quartile (Q1) = Median of the first 9 values: Q1=52+542=53 minutesQ1 = \frac{52 + 54}{2} = 53 \text{ minutes}
  • Upper Quartile (Q3) = Median of the last 9 values: Q3=66+672=66.5 minutesQ3 = \frac{66 + 67}{2} = 66.5 \text{ minutes}
  • IQR = 66.553=13.566.5 - 53 = 13.5 minutes
  • Range = 7740=3777 - 40 = 37 minutes
  • Expected IQR = 372=18.5\frac{37}{2} = 18.5 minutes, so x=0x = 0 is not correct.

Case 2: x=5x = 5 (i.e., 4545 minutes)

  • Data: 45, 47, 50, 51, 52, 54, 54, 56, 57, 59, 60, 62, 63, 66, 67, 71, 72, 73, 77
  • Median (Q2) = 57+592=58\frac{57 + 59}{2} = 58 minutes
  • Lower Quartile (Q1) = 5454 minutes
  • Upper Quartile (Q3) = 6767 minutes
  • IQR = 6754=1367 - 54 = 13 minutes
  • Range = 7745=3277 - 45 = 32 minutes
  • Expected IQR = 322=16\frac{32}{2} = 16 minutes, so x=5x = 5 is not correct.

Case 3: x=7x = 7 (i.e., 4747 minutes)

  • Data: 47, 47, 50, 51, 52, 54, 54, 56, 57, 59, 60, 62, 63, 66, 67, 71, 72, 73, 77
  • Median (Q2) = 57+592=58\frac{57 + 59}{2} = 58 minutes
  • Lower Quartile (Q1) = 5454 minutes
  • Upper Quartile (Q3) = 6767 minutes
  • IQR = 6754=1367 - 54 = 13 minutes
  • Range = 7747=3077 - 47 = 30 minutes
  • Expected IQR = 302=15\frac{30}{2} = 15 minutes, so x=7x = 7 is correct.

Answer: x=7x = 7 makes the IQR equal to 15, satisfying the condition.


Do you want more details or have any other questions? Here are some related questions to explore:

  1. How does the value of xx influence the range and quartiles of a data set?
  2. What is the importance of checking different possible values for a variable in data analysis?
  3. How would the results differ if the stem-and-leaf plot had more or fewer data points?
  4. What are some real-life examples where adjusting one data point significantly changes the statistical summary?
  5. How would you create a stem-and-leaf plot for a different data set?

Tip: When calculating quartiles, always ensure that the data is ordered correctly and that you consider all possible scenarios when a variable is involved.

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Math Problem Analysis

Mathematical Concepts

Statistics
Data Analysis
Stem-and-leaf Plot

Formulas

Interquartile Range (IQR)
Range Calculation

Theorems

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Suitable Grade Level

High School