2021 HKDSE Mathematics M1 (Calculus and Statistics) Answers & Marking Scheme

\((1 + ax)^4 = 1 + 4ax + 6a^2 x^2 + \dots\)
\((1 - x)^n = 1 - nx + \frac{n(n-1)}{2} x^2 - \dots\)
So, \((1 + ax)^4 (1 - x)^n = (1 + 4ax + 6a^2 x^2 + \dots)(1 - nx + \frac{n(n-1)}{2} x^2 - \dots)\)
The coefficient of \(x\) is \(4a - n = -1\) ...... (1)
The coefficient of \(x^2\) is \(6a^2 - 4an + \frac{n(n-1)}{2} = -12\) ...... (2)
From (1), we have \(n = 4a + 1\).
Substitute \(n = 4a + 1\) into (2):
\(6a^2 - 4a(4a + 1) + \frac{(4a+1)(4a)}{2} = -12\)
\(6a^2 - 16a^2 - 4a + 2a(4a+1) = -12\)
\(-10a^2 - 4a + 8a^2 + 2a = -12\)
\(-2a^2 - 2a + 12 = 0\)
\(a^2 + a - 6 = 0\)
\((a+3)(a-2) = 0\)
\(a = 2\) or \(a = -3\)
Since \(n\) must be a positive integer:
If \(a = -3\), then \(n = 4(-3) + 1 = -11\) (rejected).
If \(a = 2\), then \(n = 4(2) + 1 = 9\) (accepted).
Therefore, \(a = 2\) and \(n = 9\).

For setting up equation (1) for coefficient of \(x\): 1M
For setting up equation (2) for coefficient of \(x^2\): 1M
For substituting \(n = 4a + 1\) into (2): 1M
For solving the quadratic equation to find \(a = 2\) or \(a = -3\): 1.25A
For rejecting \(a = -3\) with reasons: 1M
For obtaining both final answers \(a = 2\) and \(n = 9\): 1A

(a) Let \(X\) be the number of customers arriving in a 10-minute interval. \(X \sim Po(4)\).
\(P(X = 3) = \frac{e^{-4} 4^3}{3!} = \frac{32}{3} e^{-4} \approx 0.1954\).
(b) Let \(Y\) be the number of customers arriving in a 5-minute interval. \(Y \sim Po(2)\).
\(P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1) = 1 - e^{-2} - 2e^{-2} = 1 - 3e^{-2} \approx 0.5940\).
(c) We want to find the conditional probability \(P(Y \le 4 \mid Y \ge 2)\).
\(P(2 \le Y \le 4) = P(Y = 2) + P(Y = 3) + P(Y = 4)\)
\(= \frac{e^{-2} 2^2}{2!} + \frac{e^{-2} 2^3}{3!} + \frac{e^{-2} 2^4}{4!} = e^{-2} (2 + \frac{4}{3} + \frac{2}{3}) = 4e^{-2}\).
\(P(Y \le 4 \mid Y \ge 2) = \frac{P(2 \le Y \le 4)}{P(Y \ge 2)} = \frac{4e^{-2}}{1 - 3e^{-2}} = \frac{4}{e^2 - 3} \approx 0.9114\).

(a) 1M for Poisson formula, 1A for 0.1954 (or \(\frac{32}{3} e^{-4}\)).
(b) 1M for identifying mean = 2, 1.25A for 0.5940 (or \(1 - 3e^{-2}\)).
(c) 1M for conditional probability definition, 1A for 0.9114 (or \(\frac{4}{e^2-3}\)).

(a) For 4 subintervals over \([0, 2]\), the width of each subinterval is \(h = \frac{2 - 0}{4} = 0.5\).
Let the grid points be \(x_0 = 0\), \(x_1 = 0.5\), \(x_2 = 1.0\), \(x_3 = 1.5\), and \(x_4 = 2.0\).
Applying the trapezoidal rule:
\(\int_0^2 x e^{-x} dx \approx \frac{h}{2} [f(0) + 2(f(0.5) + f(1.0) + f(1.5)) + f(2.0)]\)
\(= \frac{0.5}{2} [0 e^0 + 2(0.5 e^{-0.5} + 1.0 e^{-1} + 1.5 e^{-1.5}) + 2 e^{-2}]\)
\(\approx 0.25 [0 + 2(0.3032653 + 0.3678794 + 0.3346952) + 0.2706706]\)
\(= 0.25 [2(1.005840) + 0.2706706]\)
\(\approx 0.5706\).
(b) Differentiating \(f(x)\):
\(f'(x) = e^{-x} - x e^{-x} = (1 - x)e^{-x}\)
\(f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}\)
For \(0 < x < 2\), \(x - 2 < 0\) and \(e^{-x} > 0\), hence \(f''(x) < 0\).
Since \(f''(x) < 0\) on \((0, 2)\), the curve is concave downwards.
Therefore, the estimate obtained by the trapezoidal rule is an under-estimate.

(a) 0.5M for calculating \(h = 0.5\). 1.5M for applying the trapezoidal rule formula correctly. 0.25M for substituting the function values. 1A for the correct answer 0.5706.
(b) 1M for finding \(f'(x)\). 1M for finding \(f''(x) = (x - 2)e^{-x}\). 1A for concluding that it is an under-estimate based on \(f''(x) < 0\).

(a) Let \(X\) be the lifetime of a light bulb. \(X \sim N(\mu, \sigma^2)\).
We are given:
\(P(X < 850) = 0.0668 \implies P(Z < \frac{850 - \mu}{\sigma}) = 0.0668\)
From the standard normal distribution table, \(P(Z < -1.5) = 0.0668\), so:
\(\frac{850 - \mu}{\sigma} = -1.5 \implies \mu - 1.5\sigma = 850\) ...... (1)
Also:
\(P(X > 1120) = 0.1151 \implies P(Z > \frac{1120 - \mu}{\sigma}) = 0.1151\)
From the standard normal distribution table, \(P(Z > 1.2) = 0.1151\), so:
\(\frac{1120 - \mu}{\sigma} = 1.2 \implies \mu + 1.2\sigma = 1120\) ...... (2)
Subtract (1) from (2):
\(2.7\sigma = 270 \implies \sigma = 100\)
Substitute \(\sigma = 100\) back into (2):
\[\mu = 1120 - 1.2(100) = 1000\]
So \(\mu = 1000\) and \(\sigma = 100\).
(b) The probability that a randomly selected light bulb has a lifetime of more than 1120 hours is \(p = 0.1151\).
Let \(Y\) be the number of light bulbs with lifetimes of more than 1120 hours in a batch of 5. \(Y \sim B(5, 0.1151)\).
\(P(Y \ge 1) = 1 - P(Y = 0) = 1 - (1 - 0.1151)^5 = 1 - (0.8849)^5 \approx 0.4575\).

(a) 1.5M for setting up the equation \(\mu - 1.5\sigma = 850\) using standard normal tables. 1.5M for setting up \(\mu + 1.2\sigma = 1120\). 1A for finding both \(\mu = 1000\) and \(\sigma = 100\).
(b) 1.25M for writing down the binomial formula \(1 - (1-0.1151)^5\). 1A for the correct probability 0.4575.

To find the equation of the curve, we integrate \(\frac{dy}{dx}\):
\(y = \int x \sqrt{2x^2 + 1} dx\)
Let \(u = 2x^2 + 1\). Then \(du = 4x dx\), which means \(x dx = \frac{du}{4}\).
Substitute these into the integral:
\(y = \int \sqrt{u} \frac{du}{4} = \frac{1}{4} \int u^{1/2} du\)
\(y = \frac{1}{4} \left( \frac{2}{3} u^{3/2} \right) + C = \frac{1}{6} u^{3/2} + C\)
\(y = \frac{1}{6} (2x^2 + 1)^{3/2} + C\)
Since the curve passes through \((2, 6)\), substitute \(x = 2\) and \(y = 6\):
\(6 = \frac{1}{6} (2(2)^2 + 1)^{3/2} + C\)
\(6 = \frac{1}{6} (9)^{3/2} + C\)
\(6 = \frac{1}{6} (27) + C\)
\(6 = 4.5 + C \implies C = 1.5 = \frac{3}{2}\)
Therefore, the equation of the curve is \(y = \frac{1}{6} (2x^2 + 1)^{3/2} + \frac{3}{2}\).

For setting up the integral: 1M
For using the substitution method (such as \(u = 2x^2 + 1\)): 1.5M
For obtaining the integrated term \(\frac{1}{6} (2x^2 + 1)^{3/2}\): 1.5M
For substituting \((2, 6)\) to find the integration constant \(C = \frac{3}{2}\): 1.25M
For the correct final equation: 1A

(a) Let \(X\) be the number of defective switches in a batch of 40. \(X \sim B(40, p)\).
We have \(E(X) = np = 40p = 2.4 \implies p = 0.06\).
(b) The variance of \(X\) is:
\(Var(X) = np(1-p) = 40(0.06)(1 - 0.06) = 2.4(0.94) = 2.256\).
(c) Let \(Y\) be the number of defective switches in a batch of 15. \(Y \sim B(15, 0.06)\).
(i) \(P(Y = 1) = \binom{15}{1} (0.06)^1 (0.94)^{14} = 15(0.06)(0.42045) \approx 0.3784\).
(ii) \(P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1)\)
\(= 1 - (0.94)^{15} - 0.378405\)
\(\approx 1 - 0.395290 - 0.378405 = 0.2263\).

(a) 0.5M for setting up \(40p = 2.4\), 0.75A for \(p = 0.06\).
(b) 0.5M for setting up \(Var(X) = 40(0.06)(0.94)\), 0.5A for \(2.256\).
(c)(i) 1M for binomial term \(\binom{15}{1} (0.06)^1 (0.94)^{14}\), 1A for 0.3784.
(c)(ii) 1M for expressing \(1 - P(Y=0) - P(Y=1)\), 1A for 0.2263.

Let \(A\) be the event that a component is produced by Machine A.
Let \(B\) be the event that a component is produced by Machine B.
Let \(D\) be the event that a component is defective.
We are given:
\(P(A) = 0.60\), \(P(B) = 0.40\)
\(P(D \mid A) = 0.02\), \(P(D \mid B) = 0.05\)
(a) Using the total probability theorem:
\(P(D) = P(A) P(D \mid A) + P(B) P(D \mid B)\)
\(= (0.60)(0.02) + (0.40)(0.05)\)
\(= 0.012 + 0.020 = 0.032\).
(b) Using Bayes' theorem:
\(P(A \mid D) = \frac{P(A \cap D)}{P(D)} = \frac{P(A) P(D \mid A)}{P(D)}\)
\(= \frac{0.012}{0.032} = \frac{3}{8} = 0.375\).

(a) 1M for total probability formula, 0.5M for substitution, 0.75A for 0.032 (or \(\frac{4}{125}\)).
(b) 1.5M for Bayes' theorem formula, 1.5M for substitution, 1A for 0.375 (or \(\frac{3}{8}\)).

(a) The width of the base is \(x\) m, so the length of the base is \(2x\) m.
Let \(h\) m be the height of the container.
The volume \(V\) is given by:
\(V = x(2x)h = 2x^2 h = 36 \implies h = \frac{18}{x^2}\).
Since the container has an open top, the total surface area \(A\) is:
\(A = \text{Area of base} + 2 \times \text{Area of front/back} + 2 \times \text{Area of sides}\)
\(A = x(2x) + 2(2x)(h) + 2(x)(h) = 2x^2 + 6xh\).
Substitute \(h = \frac{18}{x^2}\) into the equation:
\(A = 2x^2 + 6x \left(\frac{18}{x^2}\right) = 2x^2 + \frac{108}{x}\).
(b) Differentiating \(A\) with respect to \(x\):
\(\frac{dA}{dx} = 4x - \frac{108}{x^2}\).
To find the critical point, set \(\frac{dA}{dx} = 0\):
\(4x - \frac{108}{x^2} = 0 \implies 4x^3 = 108 \implies x^3 = 27 \implies x = 3\).
To determine if this minimizes the area, find the second derivative:
\(\frac{d^2A}{dx^2} = 4 + \frac{216}{x^3}\).
At \(x = 3\):
\(\frac{d^2A}{dx^2} = 4 + \frac{216}{27} = 12 > 0\).
Since the second derivative is positive, \(A\) is minimized when \(x = 3\).

(a) 1M for expressing \(h = \frac{18}{x^2}\). 1M for setting up total surface area \(A = 2x^2 + 6xh\). 0.5A for completing the algebraic steps to show the formula.
(b) 1M for finding \(\frac{dA}{dx}\). 1M for setting \(\frac{dA}{dx} = 0\) and solving. 0.75A for obtaining \(x = 3\). 1M for checking that the second derivative is positive to verify the minimum.

(a) We first find the derivative \( C'(t) \):
\( C'(t) = 16t e^{-0.5t} + 8t^2 (-0.5 e^{-0.5t}) = (16t - 4t^2) e^{-0.5t} = 4t(4-t) e^{-0.5t} \).
For \( C(t) \) to be increasing, we require \( C'(t) > 0 \).
Since \( e^{-0.5t} > 0 \) and \( t > 0 \), we have:
\( 4t(4-t) > 0 \implies 0 < t < 4 \).
Therefore, the concentration is increasing for the interval \( 0 < t < 4 \).

(b) From (a), \( C'(t) > 0 \) for \( 0 < t < 4 \) and \( C'(t) < 0 \) for \( t > 4 \).
Thus, \( C(t) \) attains its absolute maximum at \( t = 4 \).
Maximum concentration is:
\( C(4) = 8(4)^2 e^{-0.5(4)} = 128 e^{-2} \approx 17.33 \).

(c) Find the second derivative:
\( C''(t) = \frac{d}{dt} \left[ (16t - 4t^2) e^{-0.5t} \right] \)
\( = (16 - 8t) e^{-0.5t} - 0.5 (16t - 4t^2) e^{-0.5t} \)
\( = (16 - 16t + 2t^2) e^{-0.5t} = 2(t^2 - 8t + 8) e^{-0.5t} \).
Setting \( C''(t) = 0 \):
\( t^2 - 8t + 8 = 0 \implies t = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2} \).
Since both \( t = 4 - 2\sqrt{2} \approx 1.17 \) and \( t = 4 + 2\sqrt{2} \approx 6.83 \) are greater than 0, they are in the domain.
Since \( C''(t) \) changes sign around these two values, both correspond to points of inflexion.
Substituting these values back into \( C(t) \):
For \( t = 4 - 2\sqrt{2} \):
\( C(4-2\sqrt{2}) = 8(4-2\sqrt{2})^2 e^{-0.5(4-2\sqrt{2})} = 8(24 - 16\sqrt{2}) e^{\sqrt{2}-2} = 64(3 - 2\sqrt{2}) e^{\sqrt{2}-2} \).
For \( t = 4 + 2\sqrt{2} \):
\( C(4+2\sqrt{2}) = 8(4+2\sqrt{2})^2 e^{-0.5(4+2\sqrt{2})} = 8(24 + 16\sqrt{2}) e^{-\sqrt{2}-2} = 64(3 + 2\sqrt{2}) e^{-\sqrt{2}-2} \).
So the points of inflexion are:
\( \left( 4 - 2\sqrt{2}, 64(3 - 2\sqrt{2}) e^{\sqrt{2}-2} \right) \) and \( \left( 4 + 2\sqrt{2}, 64(3 + 2\sqrt{2}) e^{-\sqrt{2}-2} \right) \).

(d) As \( t \to \infty \), \( C'(t) = (16t - 4t^2)e^{-0.5t} \to 0 \).
Thus, the rate of change of concentration of the drug approaches 0 (the concentration becomes stable).

(a)
- 1M: For finding the derivative \( C'(t) \)
- 1M: For setting \( C'(t) > 0 \)
- 1A: For correct interval \( 0 < t < 4 \) (or equivalent)

(b)
- 1M: For showing testing or explaining why maximum occurs at \( t = 4 \)
- 1A: For finding \( t = 4 \)
- 1A: For finding maximum value \( 128 e^{-2} \)

(c)
- 1M: For attempt to differentiate \( C'(t) \)
- 1A: For correct second derivative \( C''(t) = 2(t^2-8t+8)e^{-0.5t} \)
- 1A: For both \( t \)-coordinates of the points of inflexion
- 1M: For checking concavity change or setting \( C''(t)=0 \)
- 1A: For both exact coordinates

(d)
- 1M: For attempting the limit as \( t \to \infty \)
- 1A: For stating the rate of change approaches 0

(a) Let \( X \) be the number of complaints received in a given hour. \( X \sim \text{Po}(1.8) \).
(i) \( P(X = 2) = \frac{e^{-1.8} (1.8)^2}{2!} = 1.62 e^{-1.8} \approx 0.2678 \).
(ii) \( P(X \ge 3) = 1 - P(X \le 2) = 1 - e^{-1.8} \left(1 + 1.8 + \frac{1.8^2}{2}\right) = 1 - 4.42 e^{-1.8} \approx 0.2694 \).

(b) (i) First, find the probability \( p \) that the center receives at least 1 complaint in a given hour:
\( p = P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-1.8} \approx 0.834701 \).
Let \( W \) be the number of hours in which the center receives at least 1 complaint in a day of 8 hours.
\( W \sim B(8, p) \).
We want \( P(W \ge 6) = P(W = 6) + P(W = 7) + P(W = 8) \)
\( = \binom{8}{6} p^6 (1-p)^2 + \binom{8}{7} p^7 (1-p) + p^8 \)
\( = 28 (0.834701)^6 (0.165299)^2 + 8 (0.834701)^7 (0.165299) + (0.834701)^8 \)
\( \approx 0.259871 + 0.374944 + 0.236660 = 0.8715 \).

(ii) Let \( T \) be the total number of complaints received in an 8-hour day.
Since complaints in different hours are independent, we have \( T \sim \text{Po}(1.8 \times 8) = \text{Po}(14.4) \).
We want to find \( P(T \le 10 \mid T \ge 1) = \frac{P(1 \le T \le 10)}{P(T \ge 1)} = \frac{P(T \le 10) - P(T = 0)}{1 - P(T = 0)} \).
Now, \( P(T = 0) = e^{-14.4} \approx 5.57 \times 10^{-7} \).
\( P(T \le 10) = e^{-14.4} \sum_{k=0}^{10} \frac{14.4^k}{k!} \approx e^{-14.4} (270295.2939) \approx 0.150626 \).
Hence,
\( P(T \le 10 \mid T \ge 1) = \frac{0.150626 - 5.57 \times 10^{-7}}{1 - 5.57 \times 10^{-7}} \approx 0.1506 \).

(a)(i)
- 1M: For using formula of Poisson distribution
- 1A: For correct probability \( \approx 0.2678 \)
(a)(ii)
- 1M: For using complement probability
- 1A: For correct probability \( \approx 0.2694 \)

(b)(i)
- 1A: For correct binomial probability \( p = 1-e^{-1.8} \approx 0.8347 \)
- 1M: For setting up the binomial sum \( P(W=6) + P(W=7) + P(W=8) \)
- 1M: For substituting binomial formula correctly
- 1A: For correct probability \( \approx 0.8715 \)

(b)(ii)
- 1A: For correct mean \( 14.4 \) of Poisson distribution of \( T \)
- 1M: For expressing conditional probability formula
- 1M: For expansion of \( P(T \le 10) \)
- 1A: For correct probability \( \approx 0.1506 \)

(a) Let \( u = e^x \), then \( du = e^x dx \implies dx = \frac{du}{u} \).
When \( x = 0 \), \( u = e^0 = 1 \).
When \( x = 2 \), \( u = e^2 \).
Thus,
\( \int_{0}^{2} \frac{4}{e^x+1} dx = \int_{1}^{e^2} \frac{4}{u(u+1)} du \).
Using partial fractions:
\( \frac{4}{u(u+1)} = \frac{4}{u} - \frac{4}{u+1} \).
Therefore,
\( \int_{1}^{e^2} \left( \frac{4}{u} - \frac{4}{u+1} \right) du = 4 \left[ \ln|u| - \ln|u+1| \right]_1^{e^2} \)
\( = 4 \left( [\ln(e^2) - \ln(e^2+1)] - [\ln(1) - \ln(2)] \right) \)
\( = 4 \left( 2 - \ln(e^2+1) + \ln 2 \right) \)
\( = 8 + 4\ln 2 - 4\ln(e^2+1) \).

(b) (i) Let \( n = 4 \). The width of each sub-interval is \( \Delta x = \frac{2 - 0}{4} = 0.5 \).
The grid points are \( x_0 = 0, x_1 = 0.5, x_2 = 1.0, x_3 = 1.5, x_4 = 2.0 \).
\( f(0) = 2 \)
\( f(0.5) = \frac{4}{e^{0.5}+1} \approx 1.510125 \)
\( f(1.0) = \frac{4}{e+1} \approx 1.075766 \)
\( f(1.5) = \frac{4}{e^{1.5}+1} \approx 0.729702 \)
\( f(2.0) = \frac{4}{e^2+1} \approx 0.476812 \)
Using the trapezoidal rule:
\( \int_{0}^{2} f(x) dx \approx \frac{0.5}{2} \left[ f(0) + 2(f(0.5) + f(1.0) + f(1.5)) + f(2.0) \right] \)
\( = 0.25 \left[ 2 + 2(1.510125 + 1.075766 + 0.729702) + 0.476812 \right] \)
\( = 0.25 \left[ 2 + 2(3.315593) + 0.476812 \right] \)
\( = 0.25 \left[ 9.107998 \right] \approx 2.2770 \).

(ii) Let us find the second derivative of \( f(x) \):
\( f(x) = 4(e^x + 1)^{-1} \)
\( f'(x) = -4e^x(e^x + 1)^{-2} \)
\( f''(x) = -4e^x(e^x + 1)^{-2} + 8e^{2x}(e^x + 1)^{-3} \)
\( = \frac{-4e^x(e^x + 1) + 8e^{2x}}{(e^x+1)^3} = \frac{4e^{2x} - 4e^x}{(e^x+1)^3} = \frac{4e^x(e^x - 1)}{(e^x+1)^3} \).
Since \( e^x > 1 \) for \( 0 < x \le 2 \), we have \( f''(x) > 0 \) for all \( x \in (0, 2] \).
Since \( f''(x) \ge 0 \) on \( [0, 2] \), the curve is concave upwards.
Therefore, the estimate in (b)(i) is an over-estimate.

(c) We compute:
\( \int_{0}^{2} \frac{4}{e^x+1} dx + \int_{0}^{2} \frac{4e^x}{e^x+1} dx = \int_{0}^{2} \frac{4 + 4e^x}{e^x+1} dx \)
\( = \int_{0}^{2} \frac{4(1+e^x)}{e^x+1} dx = \int_{0}^{2} 4 dx = [4x]_0^2 = 8 \).
Hence,
\( \int_{0}^{2} \frac{4e^x}{e^x+1} dx = 8 - \int_{0}^{2} \frac{4}{e^x+1} dx \)
\( = 8 - (8 + 4\ln 2 - 4\ln(e^2+1)) \)
\( = 4\ln(e^2+1) - 4\ln 2 = 4\ln\left(\frac{e^2+1}{2}\right) \).

(a)
- 1M: For substitution and change of variable (including dx to du)
- 1A: For changing the limits of integration correctly
- 1M: For partial fractions or suitable integration technique
- 1A: For correct exact answer

(b)(i)
- 1M: For using correct trapezoidal rule formula
- 1A: For correct function values
- 1A: For correct answer \( \approx 2.2770 \) (accept 2.28)
(b)(ii)
- 1M: For correct differentiation to find \( f''(x) \)
- 1A: For establishing \( f''(x) > 0 \) on the interval
- 1A: For conclusion of over-estimate

(c)
- 1M: For putting integrals together
- 1A: For showing the sum equals 8
- 1A: For obtaining correct exact answer

(a) Let \( X \) be the weight of a package of coffee powder. We are given \( X \sim N(250, 4^2) \).
\( P(X < 245) = P\left( Z < \frac{245 - 250}{4} \right) = P(Z < -1.25) \)
\( = 0.5 - P(0 < Z < 1.25) = 0.5 - 0.3944 = 0.1056 \).

(b) We are given \( P(X < w) = 0.015 \).
Standardizing this:
\( P\left( Z < \frac{w - 250}{4} \right) = 0.015 \).
Since \( 0.015 < 0.5 \), the z-score must be negative.
Let \( z_0 = \frac{250 - w}{4} > 0 \).
Then \( P(Z > z_0) = 0.015 \implies P(0 < Z < z_0) = 0.485 \).
From the standard normal table, \( z_0 = 2.17 \).
Thus,
\( \frac{250 - w}{4} = 2.17 \implies 250 - w = 8.68 \implies w = 241.32 \approx 241.3 \).

(c) (i) Let \( p = 0.1056 \) be the probability that a package weighs less than 245 grams.
Let \( Y \) be the number of packages in a box of 6 that weigh less than 245 grams.
\( Y \sim B(6, 0.1056) \).
A box is underweight if \( Y \ge 2 \).
\( P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1) \)
\( = 1 - (1 - 0.1056)^6 - \binom{6}{1}(0.1056)^1(1 - 0.1056)^5 \)
\( = 1 - (0.8944)^6 - 6(0.1056)(0.8944)^5 \)
\( \approx 1 - 0.511790 - 0.362556 = 0.125654 \approx 0.1257 \).

(ii) We want to find the conditional probability:
\( P(Y = 1 \mid Y < 2) = \frac{P(Y = 1)}{P(Y < 2)} = \frac{P(Y = 1)}{P(Y = 0) + P(Y = 1)} \)
\( \approx \frac{0.362556}{0.511790 + 0.362556} = \frac{0.362556}{0.874346} \approx 0.414660 \approx 0.4147 \).

(a)
- 1M: For standardizing variable
- 1A: For getting \( Z < -1.25 \)
- 1A: For correct probability \( 0.1056 \)

(b)
- 1M: For setting up standardizing equation
- 1A: For finding critical z-value \( 2.17 \) (or \( -2.17 \))
- 1A: For correct value \( w = 241.3 \)

(c)(i)
- 1M: For using Binomial distribution formula with complement
- 1A: For correct substitution \( (0.8944)^6 + 6(0.1056)(0.8944)^5 \)
- 1A: For correct probability \( \approx 0.1257 \)
(c)(ii)
- 1M: For conditional probability formula \( P(Y=1)/P(Y<2) \)
- 1A: For correct numerator and denominator substitution
- 1A: For correct probability \( \approx 0.4147 \)