Solution — 2024 Question 12 Binomial Distribution| Normal Distribution

(a)

Let $A$ kWh be the amount of energy produced by a randomly chosen panel of type $A .$

Let $B$ kWh be the amount of energy produced by a randomly chosen panel of type $B .$

$A \sim N (0.255, {0.018}^{2})$

$B \sim N (0.265, {0.02}^{2})$

\begin{matrix} A - B & \sim N (0.255 - 0.265, {0.018}^{2} + {0.02}^{2}) \\ A - B & \sim N (- 0.01, 0.000 724) \end{matrix}

\begin{matrix} P (A > B) & = P (A - B > 0) \\ = 0.355 08 \\ = 0.355 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(A - B > 0\right)

(b)

Let $A_{T} = A_{1} + \dots + A_{10}$ be the total energy produced by $10$ panels of type $A .$

\begin{matrix} A_{T} & \sim N (10 (0.255), 10 {(0.018)}^{2}) \\ A_{T} & \sim N (2.55, 0.003 24) \end{matrix}

Let $B_{T} = B_{1} + \dots + B_{10}$ be the total energy produced by $10$ panels of type $B .$

\begin{matrix} B_{T} & \sim N (10 (0.265), 10 {(0.02)}^{2}) \\ B_{T} & \sim N (2.65, 0.004) \end{matrix}

\begin{matrix} B_{T} - A_{T} & \sim N (2.65 - 2.55, 0.004 + 0.003 24) \\ B_{T} - A_{T} & \sim N (0.1, 0.007 24) \end{matrix}

\begin{matrix} P (B_{T} - A_{T} > 0.15) & = 0.278 39 \\ = 0.278 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(B{T} - A{T} > 0.15\right)

(c)

Let $A_{sum} = A_{1} + \dots + A_{20}$ be the total energy produced by $20$ panels of type $A .$

\begin{matrix} A_{sum} & \sim N (20 (0.255), 20 {(0.018)}^{2}) \\ A_{sum} & \sim N (5.1, 0.006 48) \end{matrix}

Let $B_{sum} = B_{1} + \dots + B_{20}$ be the total energy produced by $20$ panels of type $B .$

\begin{matrix} B_{sum} & \sim N (20 (0.265), 20 {(0.02)}^{2}) \\ B_{sum} & \sim N (5.3, 0.008) \end{matrix}

\begin{matrix} A_{sum} + B_{sum} & \sim N (5.1 + 5.3, 0.006 48 + 0.008) \\ A_{sum} + B_{sum} & \sim N (10.4, 0.014 48) \end{matrix}

\begin{matrix} P (A_{sum} + B_{sum} > 10.5) & = 0.202 98 \\ = 0.203 ∎ (to 3 s.f.) \end{matrix}

finding \mathrm{P}\left(A{\text{sum}} + B{\text{sum}} > 10.5\right)

(d)

Let $X$ be the number of solar trees (out of 15) that produce more than 10.5 kWh.

$X \sim B (15, 0.202 98)$

\begin{matrix} P (X \geq 4) & = 1 - P (X \leq 3) \\ = 1 - 0.636 97 \\ = 0.363 03 \\ = 0.363 ∎ (to 3 s.f.) \end{matrix}

probability of X \geq 4