(a) We start with the given equation 3μ = 7σ² and the probability P(X > 2μ) = 0.1016. Standardizing, we have:

\(z = \frac{2μ - μ}{σ} = \frac{μ}{σ}\)

\(From the standard normal distribution table, P(Z > 1.272) = 0.1016, so:\)

\(\frac{7σ}{3} = 1.272\)

Solving for σ, we get:

\(σ = 0.545\)

Substituting back to find μ:

\(μ = \frac{7σ^2}{3} = 0.693\)

(b) Given Y ~ N(33, 21), we need to find a such that P(33 − a < Y < 33 + a) = 0.5. This implies:

\(P(X < a + 33) = 0.75\)

\(From the standard normal distribution table, z = 0.674 for 0.75 probability. Standardizing:\)

\(\frac{a + 33 - 33}{\sqrt{21}} = 0.674\)

Solving for a gives:

\(a = 3.09\)

(i) Given that 79% of people have visits lasting less than 10 minutes, we find the z-score corresponding to 0.79, which is 0.807. Using the formula for the z-score:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 10\), \(\mu = 8.2\), and \(z = 0.807\). Solving for \(\sigma\):

\(0.807 = \frac{10 - 8.2}{\sigma}\)

\(\sigma = \frac{1.8}{0.807} \approx 2.23\)

(ii) We need to find \(P(|X - 8.2| > 1)\), which is equivalent to \(P(X > 9.2) + P(X < 7.2)\). Standardizing these values:

\(P\left(\left| z \right| > \frac{1}{2.23}\right)\)

\(P(|z| > 0.4484) = 2 \times (1 - \Phi(0.4484))\)

\(= 2 \times (1 - 0.6729) = 0.654\)

(iii) Let \(p = 0.21\) be the probability of a visit lasting longer than 10 minutes. We want \(P(X > 2)\) for a binomial distribution with \(n = 6\) and \(p = 0.21\):

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

\(= 1 - \{ (0.79)^6 + 6 \times (0.21) \times (0.79)^5 + 15 \times (0.21)^2 \times (0.79)^4 \}\)

\(= 1 - \{ 0.2621 + 0.2371 + 0.3888 \}\)

\(= 0.112\)

(i) To find the probability that each car runs out of fuel before travelling 367 km, we standardize the normal distribution for each car.

For the Zotoc car:

\(z = \frac{367 - 320}{21.6} = 2.176\)

The probability \(P(Z < 2.176)\) is approximately 0.985.

For the Ganmor car:

\(z = \frac{367 - 350}{7.5} = 2.267\)

The probability \(P(Z < 2.267)\) is approximately 0.988.

(ii) We are given that the probability that a Zotoc car can travel at least \(320 + d\) km is 0.409. This corresponds to a z-score of 0.23 (since \(P(Z > 0.23) = 0.409\)).

Standardizing, we have:

\(0.23 = \frac{x - 320}{21.6}\)

Solving for \(x\), we get:

\(x = 0.23 \times 21.6 + 320 = 324.968\)

Thus, \(d = 324.968 - 320 = 4.97\).

(i) We have two conditions:

1. 8 out of 10 children can jump more than 127 cm, so \(P(X > 127) = 0.8\). This corresponds to a z-score of \(-0.842\).

2. 1 out of 3 children can jump more than 135 cm, so \(P(X > 135) = 0.333\). This corresponds to a z-score of \(0.431\).

Using the z-score formula \(z = \frac{x - \mu}{\sigma}\), we set up the equations:

\(0.431 = \frac{135 - \mu}{\sigma}\)

\(-0.842 = \frac{127 - \mu}{\sigma}\)

Solving these equations simultaneously gives \(\mu = 132\) and \(\sigma = 6.29\).

(ii) To find the probability that a child will not be able to jump 145 cm, we calculate \(P(X < 145)\).

Standardize: \(z = \frac{145 - 132}{6.29} = 2.023\).

Using normal distribution tables, \(P(z < 2.023) = 0.978\).

(iii) Let \(p = \frac{1}{3}\) be the probability that a child can jump more than 135 cm.

We need \(P(\text{at least 2 out of 8}) = 1 - P(0, 1)\).

\(P(0, 1) = \left(\frac{2}{3}\right)^8 + \binom{8}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^7\).

Calculate: \(P(0, 1) = 0.195\).

Thus, \(P(\text{at least 2}) = 1 - 0.195 = 0.805\).

(i) To find P(X < 2μ), we standardize the variable:

\(P(X < 2μ) = P\left( z < \frac{2μ - μ}{σ} \right) = P\left( z < \frac{μ}{σ} \right).\)

Given 5σ = 3μ, we have \(\frac{μ}{σ} = \frac{5}{3}\).

\(Thus, P(X < 2μ) = P\left( z < \frac{5}{3} \right) = 0.952.\)

(ii) We are given P(X < \(\frac{1}{3}μ\)) = 0.8524. Standardizing gives:

\(P\left( z < \frac{\frac{1}{3}μ - μ}{σ} \right) = P\left( z < \frac{-2μ}{3σ} \right).\)

From the standard normal distribution table, \(\frac{-2μ}{3σ} = -1.047\).

Solving for μ, we get \(\frac{-2μ}{3σ} = -1.047\) which gives \(μ = -1.57σ\).