We read your photo as:

The cumulative frequency table below shows the times taken by 210 students to complete a race. Find the interval for the median race time.

Mathematics

ScanSolve AI Answer

Step-by-step solution

14 steps

Here are the solutions for questions 1 to 4.

First, let's create a frequency distribution table from the given cumulative frequency table:

Time (minutes)Cumulative frequency (CF)Frequency (f)
< 2000
20 - < 301818−0=1818 - 0 = 18
30 - < 404545−18=2745 - 18 = 27
40 - < 509797−45=5297 - 45 = 52
50 - < 60163163−97=66163 - 97 = 66
60 - < 70210210−163=47210 - 163 = 47

The total number of students, N=210N = 210.

Part 1· 2 steps

For the data in Table 1, state which interval the median race time lies in. (Assuming "median weight" is a typo and should be "median race time").

  1. Find the position of the median.

    The median position is given by N2\frac{N}{2}.

    Median position=2102=105\text{Median position} = \frac{210}{2} = 105

  2. Identify the class interval containing the 105th value.

    From the cumulative frequency table:

    • The cumulative frequency for "< 50" is 97.
    • The cumulative frequency for "< 60" is 163. Since 105 is greater than 97 and less than or equal to 163, the 105th value falls into the 50 - < 60 interval.

    The median interval is C. 50 - 60.

Part 2· 2 steps

Calculate the estimate for the median for the data in Table 1.

The formula for the median of grouped data is:

Median=L+(N2−CFf)×w\text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times w

Where:

  • LL = lower boundary of the median class = 50
  • N2\frac{N}{2} = median position = 105
  • CFCF = cumulative frequency of the class before the median class = 97
  • ff = frequency of the median class (50 - < 60) = 163−97=66163 - 97 = 66
  • ww = width of the median class = 60−50=1060 - 50 = 10

  1. Substitute the values into the formula.

    Median=50+(105−9766)×10\text{Median} = 50 + \left( \frac{105 - 97}{66} \right) \times 10

  2. Simplify the expression.

    Median=50+(866)×10\text{Median} = 50 + \left( \frac{8}{66} \right) \times 10 Median=50+8066\text{Median} = 50 + \frac{80}{66} Median=50+1.2121...\text{Median} = 50 + 1.2121... Median≈51.21\text{Median} \approx 51.21

    The estimate for the median is B. 51.21.

Part 3· 5 steps

Calculate the lower quartile for the data in Table 1.

The formula for the lower quartile (Q1Q_1) of grouped data is:

Q1=L+(N4−CFf)×wQ_1 = L + \left( \frac{\frac{N}{4} - CF}{f} \right) \times w

  1. Find the position of the lower quartile.

    Q1 position=N4=2104=52.5Q_1 \text{ position} = \frac{N}{4} = \frac{210}{4} = 52.5

  2. Identify the lower quartile class.

    From the cumulative frequency table:

    • The cumulative frequency for "< 40" is 45.
    • The cumulative frequency for "< 50" is 97. Since 52.5 is greater than 45 and less than or equal to 97, the lower quartile class is 40 - < 50.

  3. Identify the values for the formula.

    • LL = lower boundary of the lower quartile class = 40
    • N4\frac{N}{4} = lower quartile position = 52.5
    • CFCF = cumulative frequency of the class before the lower quartile class = 45
    • ff = frequency of the lower quartile class (40 - < 50) = 97−45=5297 - 45 = 52
    • ww = width of the lower quartile class = 50−40=1050 - 40 = 10

  4. Substitute the values into the formula.

    Q1=40+(52.5−4552)×10Q_1 = 40 + \left( \frac{52.5 - 45}{52} \right) \times 10

  5. Simplify the expression.

    Q1=40+(7.552)×10Q_1 = 40 + \left( \frac{7.5}{52} \right) \times 10 Q1=40+7552Q_1 = 40 + \frac{75}{52} Q1=40+1.4423...Q_1 = 40 + 1.4423... Q1≈41.44Q_1 \approx 41.44

    The lower quartile is B. 41.44.

Part 4· 5 steps

Calculate the upper quartile for the data in Table 1.

The formula for the upper quartile (Q3Q_3) of grouped data is:

Q3=L+(3N4−CFf)×wQ_3 = L + \left( \frac{\frac{3N}{4} - CF}{f} \right) \times w

  1. Find the position of the upper quartile.

    Q3 position=3N4=3×2104=3×52.5=157.5Q_3 \text{ position} = \frac{3N}{4} = 3 \times \frac{210}{4} = 3 \times 52.5 = 157.5

  2. Identify the upper quartile class.

    From the cumulative frequency table:

    • The cumulative frequency for "< 50" is 97.
    • The cumulative frequency for "< 60" is 163. Since 157.5 is greater than 97 and less than or equal to 163, the upper quartile class is 50 - < 60.

  3. Identify the values for the formula.

    • LL = lower boundary of the upper quartile class = 50
    • 3N4\frac{3N}{4} = upper quartile position = 157.5
    • CFCF = cumulative frequency of the class before the upper quartile class = 97
    • ff = frequency of the upper quartile class (50 - < 60) = 163−97=66163 - 97 = 66
    • ww = width of the upper quartile class = 60−50=1060 - 50 = 10

  4. Substitute the values into the formula.

    Q3=50+(157.5−9766)×10Q_3 = 50 + \left( \frac{157.5 - 97}{66} \right) \times 10

  5. Simplify the expression.

    Q3=50+(60.566)×10Q_3 = 50 + \left( \frac{60.5}{66} \right) \times 10 Q3=50+60566Q_3 = 50 + \frac{605}{66} Q3=50+9.1666...Q_3 = 50 + 9.1666... Q3≈59.17Q_3 \approx 59.17

    The upper quartile is D. 59.17.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QThe cumulative frequency graph (ogive) shows the time taken (in minutes) for 60 employees to travel…QCalculate the median weight of students from the following frequency distribution table.QA frequency distribution table for the marks of 100 students is partially given. Complete the table…QConstruct a distribution of the following data on the length of time (minutes) if it took 80 people…QTable 2.1 gives the weights (in kg) of 35 students in a particular school. Construct a cumulative…
Browse by topic:Statistics
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.