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Normal Distribution: Independent variables — DHS 2025 H2 Math Prelim Paper 2

Dunman High School (DHS)2025 PrelimPaper 2●●○ Standard12 marks

What this question tests

Independent variables.

Question

In this question you should state the parameters of any distributions that you use.

To regulate the use of its facilities by its paying members, a gym provides a guideline of a maximum usage time of 90 minutes per visit per paying member. However, the gym's records show that the usage time, \(X\) minutes, of its paying members follows the distribution \(\mathrm{N}(95,\,\sigma^2)\), where \(\sigma\) is the standard deviation. Only 30% of its paying members adhere to the guideline.

  1. Show that \(\sigma = 9.53\), correct to 3 significant figures.

  2. The gym closes at 11 pm. John, a paying member, arrives at 9.15 pm. Find the probability that he cannot finish his workout.

  3. Ten paying members are randomly chosen. Find the probability that the usage time of the ninth paying member is the third and last one that adheres to the guideline.

To grow its membership, the gym allows each paying member to sign up a friend as a trial member. For a limited period, the trial member can use the gym for free. Over time, it is found that the usage time, \(W\) minutes, of its trial members follows the distribution \(\mathrm{N}(85,\,80)\).

  1. Sketch the distribution of \(X\) and \(W\) on the same diagram.

  2. Find the probability that the total usage time spent by two randomly chosen paying members differs from twice the time spent by a randomly chosen trial member by at least 15 minutes.

  3. State an assumption needed for your calculation in part (e) to be valid.

Show full worked solution▾

(a)

\(X \sim \mathrm{N}(95,\,\sigma^2)\), \(\mathrm{P}(X \le 90) = 0.3\).

\[\mathrm{P}\!\left(Z \le \frac{90 - 95}{\sigma}\right) = 0.3\] \[\frac{-5}{\sigma} = -0.52440\]

(b)

\(9.15\ \text{pm}\) to \(11.00\ \text{pm}\) \(= 105\) minutes.

(c)

Probability that a paying member adheres \(= \mathrm{P}(X \le 90) = 0.3\).

The 9th member is the 3rd and last to adhere: exactly 2 of the first 8 adhere, and the 9th adheres.

(d)

Figure for this question

Key features:

  • pop means 85, 95
  • diff height \(h_X < h_W\)
  • spread \(\sigma_X > \sigma_W\)
  • bell shape (asymptotic feature)
  • label \(X\) & \(W\) clearly

Key features: \(X\) has larger mean (95 vs 85) so its peak is to the right; \(\sigma_X = 9.53 > \sigma_W = \sqrt{80} \approx 8.94\) so \(X\) is slightly more spread (shorter, wider); both bell-shaped with asymptotic tails; label both curves \(X\) and \(W\); mark means 85 and 95.

(e)

Let \(W \sim \mathrm{N}(85, 80)\) (usage time of trial member).

\[\mathrm{E}(X_1 + X_2 - 2W) = 2(95) - 2(85) = 20\] \[\mathrm{Var}(X_1 + X_2 - 2W) = 2\mathrm{Var}(X) + 4\mathrm{Var}(W) = 2(9.53^2) + 4(80) = 501.64\] \[X_1 + X_2 - 2W \sim \mathrm{N}(20,\; 501.64)\] \[\begin{aligned} \mathrm{P}(|X_1 + X_2 - 2W| \ge 15) &= 1 - \mathrm{P}(-15 < X_1 + X_2 - 2W < 15)\\ &= 0.647 \text{ (3 s.f.)} \end{aligned}\]

(f)

\(X_1\), \(X_2\) and \(W\) are all mutually independent.

It is insufficient to say only "usage time of paying members are independent of trial members". The independence applies both within paying members and between paying and trial members.
Answer: (a) \(\sigma = 9.53\) (3 s.f.) (shown) (b) \(\mathrm{P}(X > 105) = 0.147\) (3 s.f.) (c) \(\mathrm{P} = 0.0623\) (3 s.f.) (e) \(\mathrm{P}(|X_1 + X_2 - 2W| \ge 15) = 0.647\) (3 s.f.) (f) \(X_1\), \(X_2\) and \(W\) are all mutually independent.
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