Let the probability of a biscuit being broken be denoted by \(p\). Since the mean number of broken biscuits is 2.7, we have:

\(18p = 2.7\)

Solving for \(p\), we get:

\(p = \frac{2.7}{18} = 0.15\)

We need to find the probability that the number of broken biscuits \(X\) is between 2 and 4 inclusive, i.e., \(P(2 \leq X \leq 4)\).

\(X\) follows a binomial distribution \(B(18, 0.15)\). Therefore,

\(P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:

\(P(X = 2) = \binom{18}{2} (0.15)^2 (0.85)^{16}\)

\(P(X = 3) = \binom{18}{3} (0.15)^3 (0.85)^{15}\)

\(P(X = 4) = \binom{18}{4} (0.15)^4 (0.85)^{14}\)

Summing these probabilities gives:

\(P(2 \leq X \leq 4) = \binom{18}{2} (0.15)^2 (0.85)^{16} + \binom{18}{3} (0.15)^3 (0.85)^{15} + \binom{18}{4} (0.15)^4 (0.85)^{14}\)

Calculating each term:

\(\binom{18}{2} (0.15)^2 (0.85)^{16} \approx 0.231\)

\(\binom{18}{3} (0.15)^3 (0.85)^{15} \approx 0.287\)

\(\binom{18}{4} (0.15)^4 (0.85)^{14} \approx 0.137\)

Adding these gives:

\(P(2 \leq X \leq 4) = 0.231 + 0.287 + 0.137 = 0.655\)

(i) We use the binomial distribution where the probability of going to the park on any day is 0.6. We need to find the probability of going to the park more than 5 times in 7 days, i.e., 6 or 7 times.

The probability of going exactly 6 times is given by:

\(\binom{7}{6} (0.6)^6 (0.4)^1 = 7 \times 0.046656 \times 0.4 = 0.1306\)

The probability of going exactly 7 times is given by:

\(\binom{7}{7} (0.6)^7 = 1 \times 0.279936 = 0.02799\)

Thus, the probability of going more than 5 times is:

\(0.1306 + 0.02799 = 0.159\)

(ii) The variance of a binomial distribution is given by \(np(1-p)\).

Here, \(n = 30\) and \(p = 0.6\).

So, the variance is:

\(30 \times 0.6 \times 0.4 = 7.2\)

Let the number of defective batteries in a pack be a binomial random variable, denoted by \(X\), with parameters \(n = 20\) and probability of defect \(p\).

Given the mean number of defective batteries is 1.6, we have:

\(20p = 1.6\)

Solving for \(p\), we get:

\(p = \frac{1.6}{20} = 0.08\)

We need to find \(P(X > 2)\). Using the complement rule:

\(P(X > 2) = 1 - P(X \leq 2)\)

Calculate \(P(X \leq 2)\) using the binomial probability formula:

\(P(X = 0) = \binom{20}{0} (0.08)^0 (0.92)^{20} = (0.92)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.08)^1 (0.92)^{19} = 20 \times 0.08 \times (0.92)^{19}\)

\(P(X = 2) = \binom{20}{2} (0.08)^2 (0.92)^{18} = \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)

Thus,

\(P(X \leq 2) = (0.92)^{20} + 20 \times 0.08 \times (0.92)^{19} + \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)

Substituting the values:

\(P(X \leq 2) = 0.1887 + 0.3281 + 0.2711 = 0.7879\)

Therefore,

\(P(X > 2) = 1 - 0.7879 = 0.212\)

The probability that a customer rates the logo as good is 0.42. Therefore, the probability that a customer does not rate the logo as good is 0.58.

The probability that none of the n customers rate the logo as good is given by:

\((0.58)^n\)

We want the probability that at least one person rates the logo as good to be greater than 0.995. Therefore, we have:

\(1 - (0.58)^n > 0.995\)

Rearranging gives:

\((0.58)^n < 0.005\)

Taking logarithms on both sides:

\(n \log(0.58) < \log(0.005)\)

Solving for n gives:

\(n > \frac{\log(0.005)}{\log(0.58)}\)

Calculating the right-hand side:

\(n > 9.727\)

Since n must be an integer, the smallest value of n is 10.

The probability that a water pistol works properly is 92% or 0.92.

The probability that all n water pistols work properly is given by:

\(P(0) = (0.92)^n\)

We want the probability that at least one does not work properly to be greater than 0.9:

\(1 - P(0) > 0.9\)

Substituting for \(P(0)\):

\(1 - (0.92)^n > 0.9\)

Rearranging gives:

\((0.92)^n < 0.1\)

Taking logarithms on both sides:

\(n \log(0.92) < \log(0.1)\)

Solving for \(n\):

\(n > \frac{\log(0.1)}{\log(0.92)}\)

Calculating gives:

\(n > 27.6\)

Since \(n\) must be an integer, the smallest possible value is:

28