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Write down the modal class of the data. Calculate the estimated mean weight-loss of the race car drivers. Sketch the ogive (cumulative frequency graph) representing race 2 in the ANSWER BOOK. How will the range of race 2 compare with the range of race 1? Determine the average weight lost in race 2.

Mathematics

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Here's the solution to Question 2:

2.1.1 Write down the modal class of the data.

  1. Identify the class interval with the highest frequency.

    From the table, the highest frequency is 15, which corresponds to the interval 2500≤w<30002500 \le w < 3000.
    The modal class is 2500 \lew < 3000\boxed{\text{2500 \le w < 3000}}.

    2.1.2 Calculate the estimated mean weight-loss of the race car drivers.

  2. Calculate the midpoint (xmx_m) for each interval and the product of frequency (ff) and midpoint (xmx_m).

    The total number of drivers is ∑f=40\sum f = 40.

    Interval of Weight Loss ww (in grams)Number of Drivers (ff)Midpoint of Interval (xmx_m)f⋅xmf \cdot x_m
    0≤w<5000 \le w < 50010+5002=250\frac{0+500}{2} = 2501×250=2501 \times 250 = 250
    500≤w<1000500 \le w < 10002500+10002=750\frac{500+1000}{2} = 7502×750=15002 \times 750 = 1500
    1000≤w<15001000 \le w < 150031000+15002=1250\frac{1000+1500}{2} = 12503×1250=37503 \times 1250 = 3750
    1500≤w<20001500 \le w < 200081500+20002=1750\frac{1500+2000}{2} = 17508×1750=140008 \times 1750 = 14000
    2000≤w<25002000 \le w < 250062000+25002=2250\frac{2000+2500}{2} = 22506×2250=135006 \times 2250 = 13500
    2500≤w<30002500 \le w < 3000152500+30002=2750\frac{2500+3000}{2} = 275015×2750=4125015 \times 2750 = 41250
    3000≤w<35003000 \le w < 350053000+35002=3250\frac{3000+3500}{2} = 32505×3250=162505 \times 3250 = 16250
    Total4090500
  3. Calculate the estimated mean using the formula xˉ=∑(f⋅xm)∑f\bar{x} = \frac{\sum (f \cdot x_m)}{\sum f}.

    xˉ=9050040=2262.5\bar{x} = \frac{90500}{40} = 2262.5

    The estimated mean weight-loss is 2262.5 grams\boxed{\text{2262.5 grams}}.

    2.2.1 Sketch the ogive (cumulative frequency graph) representing race 2 in the ANSWER BOOK.

  4. Determine the cumulative frequencies for Race 1.

    Interval of Weight Loss ww (in grams)Number of Drivers (ff)Upper BoundaryCumulative Frequency (CF)
    0≤w<5000 \le w < 50015001
    500≤w<1000500 \le w < 1000210001+2=31+2=3
    1000≤w<15001000 \le w < 1500315003+3=63+3=6
    1500≤w<20001500 \le w < 2000820006+8=146+8=14
    2000≤w<25002000 \le w < 25006250014+6=2014+6=20
    2500≤w<30002500 \le w < 300015300020+15=3520+15=35
    3000≤w<35003000 \le w < 35005350035+5=4035+5=40
  5. Determine the value of kk.

    The problem states that the amount of weight lost in race 2 was kk grams more than in race 1.
    For Race 1, the minimum weight loss is 0 grams. For Race 2, the ogive is grounded at (4;0)(4; 0), meaning the minimum weight loss is 4 grams.
    So, 0+k=4  ⟹  k=40 + k = 4 \implies k = 4.

  6. Determine the points for the ogive of Race 2 by adding k=4k=4 to the upper boundaries of Race 1.

    The ogive for Race 2 will be plotted using the following points:

    • Start: (4;0)(4; 0) (given)
    • (500+4;1)=(504;1)(500+4; 1) = (504; 1)
    • (1000+4;3)=(1004;3)(1000+4; 3) = (1004; 3)
    • (1500+4;6)=(1504;6)(1500+4; 6) = (1504; 6)
    • (2000+4;14)=(2004;14)(2000+4; 14) = (2004; 14)
    • (2500+4;20)=(2504;20)(2500+4; 20) = (2504; 20)
    • (3000+4;35)=(3004;35)(3000+4; 35) = (3004; 35)
    • End: (3500+4;40)=(3504;40)(3500+4; 40) = (3504; 40) (given)

    To sketch the ogive:

    • Plot the points (4;0)(4; 0), (504;1)(504; 1), (1004;3)(1004; 3), (1504;6)(1504; 6), (2004;14)(2004; 14), (2504;20)(2504; 20), (3004;35)(3004; 35), and (3504;40)(3504; 40).
    • Connect these points with a smooth curve. The x-axis represents weight loss (in grams) and the y-axis represents cumulative frequency (number of drivers).

    2.2.2 How will the range of race 2 compare with the range of race 1?

  7. Calculate the range for Race 1.

    The minimum weight loss for Race 1 is 0 grams. The maximum weight loss for Race 1 is the upper boundary of the last interval, which is 3500 grams.

    Range1=Maximum value−Minimum value=3500−0=3500 grams\text{Range}_1 = \text{Maximum value} - \text{Minimum value} = 3500 - 0 = 3500 \text{ grams}
  8. Calculate the range for Race 2.

    The minimum weight loss for Race 2 is 4 grams (given as grounded at (4;0)(4; 0)). The maximum weight loss for Race 2 is 3504 grams (given as the maximum value of the ogive (3504;40)(3504; 40)).

    Range2=Maximum value−Minimum value=3504−4=3500 grams\text{Range}_2 = \text{Maximum value} - \text{Minimum value} = 3504 - 4 = 3500 \text{ grams}
  9. Compare the ranges.

    The range of Race 2 is the same as the range of Race 1\boxed{\text{the same as the range of Race 1}}.

    2.2.3 Determine the average weight lost in race 2.

  10. Recall the estimated mean weight loss for Race 1 from question 2.1.2.

    Estimated mean for Race 1 (xˉ1\bar{x}_1) = 2262.5 grams.

  11. Use the relationship between the weight loss in Race 1 and Race 2.

    The amount of weight lost in Race 2 was k=4k=4 grams more than in Race 1. When all data points are shifted by a constant value, the mean also shifts by that same constant value.

    xˉ2=xˉ1+k\bar{x}_2 = \bar{x}_1 + k xˉ2=2262.5+4=2266.5\bar{x}_2 = 2262.5 + 4 = 2266.5

    The average weight lost in Race 2 is 2266.5 grams\boxed{\text{2266.5 grams}}.

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