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Draw an ogive (cumulative frequency curve) on the DIAGRAM SHEET to represent the above data

Mathematics

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Answers

7 parts
5.1

7.4\text{7.4}

5.2

7\text{7}

5.3

2.36\text{2.36}

5.4

17\text{17}

6.2

40≤t<60\text{40} \le t < \text{60}

6.3

8\text{8}

6.4

46.51 hours\text{46.51 hours}

Step-by-step solution

20 steps

Here are the solutions for Question 5 and Question 6.

Part 5

  • Data Table Setup: We first set up a table to calculate ∑xf\sum xf and ∑x2f\sum x^2f. Sum (x)Frequency (f)xfx2x2f200403399274281632542025100642436144785649392832464192921881162102201002001111112112112112144144Total∑f=30∑xf=222∑x2f=1514\begin{array}{|c|c|c|c|c|} \hline \text{Sum (x)} & \text{Frequency (f)} & xf & x^2 & x^2f \\ \hline 2 & 0 & 0 & 4 & 0 \\ 3 & 3 & 9 & 9 & 27 \\ 4 & 2 & 8 & 16 & 32 \\ 5 & 4 & 20 & 25 & 100 \\ 6 & 4 & 24 & 36 & 144 \\ 7 & 8 & 56 & 49 & 392 \\ 8 & 3 & 24 & 64 & 192 \\ 9 & 2 & 18 & 81 & 162 \\ 10 & 2 & 20 & 100 & 200 \\ 11 & 1 & 11 & 121 & 121 \\ 12 & 1 & 12 & 144 & 144 \\ \hline \text{Total} & \sum f = 30 & \sum xf = 222 & & \sum x^2f = 1514 \\ \hline \end{array}

Part 5.1

7.4\text{7.4}

Calculate the mean of the data

  1. Use the formula for the mean of grouped data.

    xˉ=∑xf∑f\bar{x} = \frac{\sum xf}{\sum f}

  2. Substitute the values from the table.

    xˉ=22230\bar{x} = \frac{222}{30}

  3. Calculate the mean.

    xˉ=7.4\bar{x} = 7.4

    The mean of the data is 7.4\boxed{\text{7.4}}.

Part 5.2

7\text{7}

Determine the median of the data

  1. Find the position of the median.

    The total number of data points is N=30N = 30. The median position is the average of the N2\frac{N}{2}-th and (N2+1)(\frac{N}{2} + 1)-th values.

    Median position=302=15-th value and 302+1=16-th value\text{Median position} = \frac{30}{2} = 15\text{-th value and } \frac{30}{2} + 1 = 16\text{-th value}

    So, the median is the average of the 15th and 16th values.

  2. Create a cumulative frequency column to locate the median values.

    Sum (x)Frequency (f)Cumulative Frequency (CF)2003334255496413782183249226102281112912130\begin{array}{|c|c|c|} \hline \text{Sum (x)} & \text{Frequency (f)} & \text{Cumulative Frequency (CF)} \\ \hline 2 & 0 & 0 \\ 3 & 3 & 3 \\ 4 & 2 & 5 \\ 5 & 4 & 9 \\ 6 & 4 & 13 \\ 7 & 8 & 21 \\ 8 & 3 & 24 \\ 9 & 2 & 26 \\ 10 & 2 & 28 \\ 11 & 1 & 29 \\ 12 & 1 & 30 \\ \hline \end{array}

    The 15th value falls in the '7' category (since CF up to 6 is 13, and CF up to 7 is 21).
    The 16th value also falls in the '7' category.

  3. Calculate the median.

    Median=7+72=7\text{Median} = \frac{7 + 7}{2} = 7

    The median of the data is 7\boxed{\text{7}}.

Part 5.3

2.36\text{2.36}

Determine the standard deviation of the data

  1. Use the formula for the population standard deviation for grouped data.

    σ=∑(x−xˉ)2f∑f\sigma = \sqrt{\frac{\sum (x - \bar{x})^2 f}{\sum f}}

    We need to calculate ∑(x−xˉ)2f\sum (x - \bar{x})^2 f.

    Sum (x)Frequency (f)x−xˉ(x−xˉ)2(x−xˉ)2f20−5.429.16033−4.419.3658.0842−3.411.5623.1254−2.45.7623.0464−1.41.967.8478−0.40.161.28830.60.361.08921.62.565.121022.66.7613.521113.612.9612.961214.621.1621.16Total30∑(x−xˉ)2f=167.2\begin{array}{|c|c|c|c|c|} \hline \text{Sum (x)} & \text{Frequency (f)} & x - \bar{x} & (x - \bar{x})^2 & (x - \bar{x})^2 f \\ \hline 2 & 0 & -5.4 & 29.16 & 0 \\ 3 & 3 & -4.4 & 19.36 & 58.08 \\ 4 & 2 & -3.4 & 11.56 & 23.12 \\ 5 & 4 & -2.4 & 5.76 & 23.04 \\ 6 & 4 & -1.4 & 1.96 & 7.84 \\ 7 & 8 & -0.4 & 0.16 & 1.28 \\ 8 & 3 & 0.6 & 0.36 & 1.08 \\ 9 & 2 & 1.6 & 2.56 & 5.12 \\ 10 & 2 & 2.6 & 6.76 & 13.52 \\ 11 & 1 & 3.6 & 12.96 & 12.96 \\ 12 & 1 & 4.6 & 21.16 & 21.16 \\ \hline \text{Total} & 30 & & & \sum (x - \bar{x})^2 f = 167.2 \\ \hline \end{array}

  2. Substitute the sum into the standard deviation formula.

    σ=167.230\sigma = \sqrt{\frac{167.2}{30}}

  3. Calculate the standard deviation.

    σ=5.57333...≈2.3608\sigma = \sqrt{5.57333...} \approx 2.3608

    The standard deviation of the data is approximately 2.36\boxed{\text{2.36}}.

Part 5.4

17\text{17}

Determine the number of times that the sum of recorded values of the dice is within ONE standard deviation from the mean. Show your calculations.

  1. Calculate the range within one standard deviation from the mean.

    Range=(xˉ−σ,xˉ+σ)\text{Range} = (\bar{x} - \sigma, \bar{x} + \sigma) Range=(7.4−2.3608,7.4+2.3608)\text{Range} = (7.4 - 2.3608, 7.4 + 2.3608) Range=(5.0392,9.7608)\text{Range} = (5.0392, 9.7608)

  2. Identify the sum values (xx) that fall within this range.

    The integer sum values within the range $(5.0392, 9.7608)$ are 6, 7, 8, and 9.

  3. Sum the frequencies for these values.

    • For x=6x=6, frequency is 4.
    • For x=7x=7, frequency is 8.
    • For x=8x=8, frequency is 3.
    • For x=9x=9, frequency is 2.
    Total frequency=4+8+3+2=17\text{Total frequency} = 4 + 8 + 3 + 2 = 17

    The number of times the sum is within one standard deviation from the mean is 17\boxed{\text{17}}.

Part 6

  • Data Table Setup: First, we convert the cumulative frequencies to individual frequencies and find the midpoints for each class. Time (hours)Cumulative Frequency (CF)Frequency (f)Midpoint (m)mf0≤t<2025251025020≤t<406969−25=4430132040≤t<60129129−69=6050300060≤t<80157157−129=2870196080≤t<100166166−157=990810100≤t<120172172−166=6110660Total∑f=172∑mf=8000\begin{array}{|c|c|c|c|c|} \hline \text{Time (hours)} & \text{Cumulative Frequency (CF)} & \text{Frequency (f)} & \text{Midpoint (m)} & mf \\ \hline 0 \le t < 20 & 25 & 25 & 10 & 250 \\ 20 \le t < 40 & 69 & 69 - 25 = 44 & 30 & 1320 \\ 40 \le t < 60 & 129 & 129 - 69 = 60 & 50 & 3000 \\ 60 \le t < 80 & 157 & 157 - 129 = 28 & 70 & 1960 \\ 80 \le t < 100 & 166 & 166 - 157 = 9 & 90 & 810 \\ 100 \le t < 120 & 172 & 172 - 166 = 6 & 110 & 660 \\ \hline \text{Total} & & \sum f = 172 & & \sum mf = 8000 \\ \hline \end{array}

Part 6.1

Draw an ogive (cumulative frequency curve) on the DIAGRAM SHEET to represent the above data

To draw an ogive:

  • Plot the upper class boundaries against their corresponding cumulative frequencies.
  • The points to plot are: $(0, 0)$, $(20, 25)$, $(40, 69)$, $(60, 129)$, $(80, 157)$, $(100, 166)$, $(120, 172)$.
  • Join these points with a smooth curve. (As an AI, I cannot physically draw the ogive, but these are the steps you would follow on a diagram sheet.)

Part 6.2

40≤t<60\text{40} \le t < \text{60}

Write down the modal class of the data

  1. Identify the class with the highest frequency from the frequency column calculated above.

    The frequencies are 25, 44, 60, 28, 9, 6.
    The highest frequency is 60.

  2. State the class interval corresponding to the highest frequency.

    The class interval corresponding to a frequency of 60 is 40≤t<6040 \le t < 60.
    The modal class of the data is 40≤t<60\boxed{\text{40} \le t < \text{60}}.

Part 6.3

8\text{8}

Use the ogive (cumulative frequency curve) to estimate the number of learners who watched television more than 80% of the time

  1. Determine what "80% of the time" refers to.

    The maximum time in the data is 120 hours.

    80% of 120 hours=0.80×120=96 hours80\% \text{ of } 120 \text{ hours} = 0.80 \times 120 = 96 \text{ hours}

    We need to estimate the number of learners who watched more than 96 hours.

  2. Estimate the cumulative frequency at 96 hours using interpolation (as an ogive is not available).

    The value 96 hours falls in the class 80≤t<10080 \le t < 100.
    Cumulative frequency at t=80t=80 is 157.
    Cumulative frequency at t=100t=100 is 166.
    The interval width is 100−80=20100 - 80 = 20. The CF increase is 166−157=9166 - 157 = 9.

    CF at 96=157+(96−80100−80)×(166−157)\text{CF at } 96 = 157 + \left(\frac{96 - 80}{100 - 80}\right) \times (166 - 157) CF at 96=157+(1620)×9\text{CF at } 96 = 157 + \left(\frac{16}{20}\right) \times 9 CF at 96=157+0.8×9\text{CF at } 96 = 157 + 0.8 \times 9 CF at 96=157+7.2=164.2\text{CF at } 96 = 157 + 7.2 = 164.2

    So, approximately 164 learners watched 96 hours or less.

  3. Calculate the number of learners who watched more than 96 hours.

    Total number of learners is 172.

    Learners watching >96 hours=Total learners−CF at 96 hours\text{Learners watching } > 96 \text{ hours} = \text{Total learners} - \text{CF at } 96 \text{ hours} Learners watching >96 hours=172−164.2=7.8\text{Learners watching } > 96 \text{ hours} = 172 - 164.2 = 7.8

    Since the number of learners must be an integer, we round to the nearest whole number.
    The estimated number of learners is 8\boxed{\text{8}}.

Part 6.4

46.51 hours\text{46.51 hours}

Estimate the mean time (in hours) that learners spent watching television during 3 weeks of the holiday

  1. Use the formula for the mean of grouped data.

    xˉ=∑mf∑f\bar{x} = \frac{\sum mf}{\sum f}

    From the table setup, ∑mf=8000\sum mf = 8000 and ∑f=172\sum f = 172.

  2. Substitute the values.

    xˉ=8000172\bar{x} = \frac{8000}{172}

  3. Calculate the mean.

    xˉ≈46.5116\bar{x} \approx 46.5116

    The estimated mean time is approximately 46.51 hours\boxed{\text{46.51 hours}}.

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