Good Papa Caf'e sells only two types of muffins, chocolate and banana. The number of chocolate and the number of banana muffins sold in a randomly chosen hour have independent Poisson distributions with means 3 and $\lambda$ respectively. Given that in a particular hour, the probability of selling 2 muffins is the same as the probability of selling 4 banana muffins, find the value of $\lambda$. If the caf'e opens for 8 hours a day, use appropriate approximations to find the probability that the daily sale of chocolate muffins exceeds that of banana muffins. Steven, who loves soccer, trains his shooting skills by trying to kick balls into the net from distances varying from 15 metres to 35 metres. The probability, $p$, that he kicks the ball into the net is given by $p = \dfrac{2}{65}(40 - d)$, where $d$ is the distance between the ball and the goalpost in metres. Each kick by him is assumed to be independent of previous kicks. Find the maximum value of $d$ (to the nearest integer) for there to be more than 60% chance of Steven kicking the ball into the net. Steven kicks 15 balls from a fixed distance of $x$ metres from the goalpost. Find, to the nearest integer, the value of $x$ such that Steven has 90% chance of kicking at least two balls into the net. Steven trains for 60 consecutive days by kicking 100 balls per day at a distance of 24 metres from the goalpost. His target is to kick an average of 50 balls into the net per day. Find the probability that the average number of balls that he kicks into the net per day over this period differs from his target by less than 1.

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,
$\text{P}(49 < \bar{X} < 51) \approx 0.63654 \approx 0.637 \text{ (to 3 s.f.)}$