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ACSEE (Form 6) Mathematics — practice by topic

Browse 10 questions across 2 Mathematics topics. Choose what you want to revise, then open a question to check its answer and worked solution.

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Mathematics topics

Statistics and Probability (9)Differential Equations (1)

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With $N=36$, $\sum fd^2=197500$ and $\sum fd=-750$, the variance of the distribution to three decimal places is:Read the answer and worked solution →With $N=36$, $L=300$, $n_b=32$, $n_w=4$ and $c=50$, the 90th percentile of the distribution is:Read the answer and worked solution →For grouped data with $L=39.5$, $N=180$, $f_b=37$, $n_w=73$ and $c=10$, the lower quartile is:Read the answer and worked solution →

Worked practice questions

With $N=36$, $\sum fd^2=197500$ and $\sum fd=-750$, the variance of the distribution to three decimal places is:

  1. 5486.111
  2. 5052.083
  3. 4618.056
  4. 434.028
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Correct answer: B. 5052.083

1. Use $\sigma^2=\dfrac{\sum fd^2}{N}-\left(\dfrac{\sum fd}{N}\right)^2$. 2. First term: $\dfrac{197500}{36}=5486.111$. 3. Second term: $\left(\dfrac{-750}{36}\right)^2=(-20.8333)^2=434.028$. 4. Subtract: $5486.111-434.028=5052.083$. 5. $5486.111$ stops after step 2; $434.028$ reports only the correction; $4618.056$ subtracts twice. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 4: Statistics_

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With $N=36$, $L=300$, $n_b=32$, $n_w=4$ and $c=50$, the 90th percentile of the distribution is:

  1. 300
  2. 310
  3. 305
  4. 32.4
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Correct answer: C. 305

1. Use $P_k=L+\left(\dfrac{k\frac{N}{100}-n_b}{n_w}\right)c$ with $k=90$. 2. Position: $90\times\dfrac{36}{100}=32.4$. 3. Numerator: $32.4-32=0.4$. 4. $P_{90}=300+\dfrac{0.4}{4}\times 50=300+5=305$. 5. Reporting $32.4$ stops at the position; adding rather than subtracting $n_b$ pushes the answer past $310$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 4: Statistics_

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For grouped data with $L=39.5$, $N=180$, $f_b=37$, $n_w=73$ and $c=10$, the lower quartile is:

  1. 40.596
  2. 39.500
  3. 45.000
  4. 44.096
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Correct answer: A. 40.596

1. Use $Q_1=L+\left(\dfrac{\frac{N}{4}-f_b}{n_w}\right)c$. 2. First find the quartile position: $\dfrac{N}{4}=\dfrac{180}{4}=45$. 3. Subtract the cumulative frequency below the class: $45-37=8$. 4. Then $\dfrac{8}{73}\times10=1.096$. 5. Adding to the lower boundary: $39.5+1.096=40.596$. Reporting 45 stops at the position rather than converting it to a value. _Source: NECTA ACSEE 2022 Advanced Mathematics 142/1, Question 4: Statistics_

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