Good Papa 咖啡馆只出售两种玛芬蛋糕：巧克力和香蕉。在随机选择的一小时内，售出的巧克力玛芬数量和香蕉玛芬数量分别服从均值为 3 和 $\lambda$ 的独立泊松分布。 已知在特定的一小时内，售出 2 个玛芬蛋糕的概率与售出 4 个香蕉玛芬蛋糕的概率相同，求 $\lambda$ 的值。 如果咖啡馆每天营业 8 小时，请使用适当的近似方法求出每天售出的巧克力玛芬数量超过香蕉玛芬数量的概率。 热爱足球的 Steven 通过尝试从 15 米到 35 米不等的距离将球踢入球网来训练他的射门技巧。他将球踢入球网的概率 $p$ 由 $p = \dfrac{2}{65}(40 - d)$ 给出，其中 $d$ 是球与球门柱之间的距离（以米为单位）。 假设他的每一次踢球都与之前的踢球相互独立。 若要使 Steven 将球踢入球网的概率大于 60%，求 $d$ 的最大值（精确到最接近的整数）。 Steven 从距离球门柱固定距离 $x$ 米处踢 15 个球。求 $x$ 的值（精确到最接近的整数），使得 Steven 有 90% 的概率将至少两个球踢入球网。 Steven 连续 60 天进行训练，每天在距离球门柱 24 米处踢 100 个球。他的目标是平均每天踢入球网 50 个球。求在此期间，他平均每天踢入球网的球数与他的目标相差小于 1 的概率。

1. Set up the probability equation for $\lambda$.

Let $T$ be the total number of muffins sold in an hour.
Since the sales of chocolate ($C$) and banana ($B$) muffins are independent, $T = C + B \sim \text{Po}(3 + \lambda)$.
Given $P(T = 2) = P(B = 4)$:
$\frac{e^{-(3+\lambda)}(3+\lambda)^2}{2!} = \frac{e^{-\lambda}\lambda^4}{4!}$

2. Solve the equation for $\lambda$.

Multiply both sides by $24e^{\lambda}$:
$12e^{-3}(3+\lambda)^2 = \lambda^4$
Taking the square root (since $\lambda > 0$):
$\sqrt{12e^{-3}}(3+\lambda) = \lambda^2$
$\lambda^2 - 0.77295\lambda - 2.31884 = 0$
Using the quadratic formula:
$\lambda = \frac{0.77295 + \sqrt{(-0.77295)^2 - 4(1)(-2.31884)}}{2}$
$\lambda \approx 1.9575$
$\therefore \lambda = 1.96 \text{ (3 s.f.)}$

3. Define variables for the 8-hour daily sales and state their Normal approximations.

Let $C_8$ and $B_8$ be the daily sales over 8 hours.
$C_8 \sim \text{Po}(8 \times 3) \Rightarrow C_8 \sim \text{Po}(24)$
$B_8 \sim \text{Po}(8 \times 1.9575) \Rightarrow B_8 \sim \text{Po}(15.660)$
Since the means are suitably large ($>10$), we apply Normal approximations:
$C_8 \sim \text{N}(24, 24)$
$B_8 \sim \text{N}(15.660, 15.660)$

4. Find the distribution of the difference $C_8 - B_8$.

Let $D = C_8 - B_8$.
$\text{E}(D) = 24 - 15.660 = 8.340$
$\text{Var}(D) = 24 + 15.660 = 39.660$
$D \sim \text{N}(8.340, 39.660)$

5. Apply the continuity correction and calculate the required probability.

We require the probability that the daily sale of chocolate muffins exceeds banana muffins, which is $P(C_8 > B_8)$ or $P(D \ge 1)$.
Using the continuity correction, we evaluate $P(D_{\text{norm}} > 0.5)$:
$P(D_{\text{norm}} > 0.5) = P\left( Z > \frac{0.5 - 8.3398}{\sqrt{39.660}} \right)$
$= P(Z > -1.2449)$
$\approx 0.893 \text{ (3 s.f.)}$

1. Set up the inequality for the probability to be more than 60%.

$\frac{2}{65}(40 - d) > 0.6$

2. Solve for $d$ and find the maximum integer value.

$40 - d > 0.6 \times \frac{65}{2}$
$40 - d > 19.5$
$d < 20.5$
Since $d$ must be strictly less than $20.5$, the maximum possible value to the nearest integer is $20$.

1. Define the random variable and set up the probability equation.

Let $Y$ be the number of balls kicked into the net out of 15.
Since each kick is independent, $Y \sim \mathrm{B}(15, p)$, where $p = \frac{2}{65}(40-x)$.
Given that the chance of kicking at least two balls is 90%,
$P(Y \ge 2) = 0.90$
$1 - P(Y \le 1) = 0.90$
$P(Y \le 1) = 0.10$

2. Express the required probability in terms of $p$.

$P(Y = 0) + P(Y = 1) = 0.10$
$(1-p)^{15} + \binom{15}{1}p(1-p)^{14} = 0.10$
$(1-p)^{15} + 15p(1-p)^{14} = 0.10$

3. Solve for $p$ using a graphing calculator.

Using a graphing calculator to solve $(1-p)^{15} + 15p(1-p)^{14} - 0.10 = 0$:
$p \approx 0.23538$

4. Calculate $x$ and round to the nearest integer.

Substitute $p \approx 0.23538$ into $p = \frac{2}{65}(40-x)$:
$\frac{2}{65}(40-x) = 0.23538$
$40-x \approx 7.650$
$x \approx 32.350$
Rounding to the nearest integer, we get $x = 32$.

1. Find the probability of a successful kick at a distance of 24 metres.

Given $p = \dfrac{2}{65}(40 - d)$.
When $d = 24$,
$p = \dfrac{2}{65}(40 - 24) = \dfrac{2}{65}(16) = \dfrac{32}{65}$

2. Define the distribution for the number of successful kicks per day and find its mean and variance.

Let $X$ be the number of balls kicked into the net in one day.
$X \sim \text{B}\left(100, \dfrac{32}{65}\right)$
$\text{E}(X) = np = 100 \left(\dfrac{32}{65}\right) = \dfrac{640}{13} \approx 49.231$
$\text{Var}(X) = np(1-p) = 100 \left(\dfrac{32}{65}\right) \left(1 - \dfrac{32}{65}\right) = \dfrac{4224}{169} \approx 24.994$

3. Determine the distribution of the sample mean $\bar{X}$ using the Central Limit Theorem.

Let $\bar{X}$ be the average number of successful kicks per day over $n=60$ days.
Since $n = 60$ is large ($n \ge 30$), by the Central Limit Theorem, $\bar{X}$ is approximately normally distributed.
$\bar{X} \sim \text{N}\left(\text{E}(X), \dfrac{\text{Var}(X)}{n}\right)$
$\bar{X} \sim \text{N}\left(\dfrac{640}{13}, \dfrac{4224/169}{60}\right)$
$\bar{X} \sim \text{N}\left(\dfrac{640}{13}, \dfrac{352}{845}\right)$

4. Calculate the probability that the sample mean differs from the target by less than 1.

His target average is 50 balls per day.
We require $\text{P}(|\bar{X} - 50| < 1)$.
$\text{P}(|\bar{X} - 50| < 1) = \text{P}(49 < \bar{X} < 51)$
Using a graphing calculator,
$\text{P}(49 < \bar{X} < 51) \approx 0.63654 \approx 0.637 \text{ (to 3 s.f.)}$