(a) We know that \(P(82-q < X < 82+q) = 0.44\). This implies \(P(X < 82+q) - P(X < 82-q) = 0.44\). Since the distribution is symmetric, \(P(X < 82+q) = 0.72\) and \(P(X < 82-q) = 0.28\). The standard normal variable \(Z\) corresponding to \(X\) is \(Z = \frac{X - 82}{7.4}\). For \(P(X < 82+q) = 0.72\), \(Z = 0.583\). Therefore, \(\frac{q}{7.4} = 0.583\) which gives \(q = 4.31\).

(b) We have \(5\mu = 2\sigma^2\) and \(P(Y < \frac{1}{2}\mu) = 0.281\). Standardizing, \(Z = \frac{\frac{1}{2}\mu - \mu}{\sigma} = \frac{-0.5\mu}{\sigma}\). From the standard normal table, \(Z = -0.580\). Thus, \(\frac{-0.5\mu}{\sigma} = -0.580\) gives \(\sigma = \frac{0.5\mu}{0.580}\). Substituting \(\sigma\) in \(5\mu = 2\sigma^2\), we solve to find \(\mu = 3.36\) and \(\sigma = 2.90\).

(i) Let \(X\) be the amount of fibre in a packet, \(X \sim N(160, \sigma^2)\). We know \(P(X > 190) = 0.19\). The z-score for 190 is given by:

\(z = \frac{190 - 160}{\sigma}\)

From the standard normal distribution table, \(P(Z > z) = 0.19\) corresponds to \(z = 0.878\).

Thus, \(\frac{30}{\sigma} = 0.878\).

Solving for \(\sigma\), we get:

\(\sigma = \frac{30}{0.878} = 34.2\)

(ii) Let \(Y\) be the number of packets with more than 190 grams of fibre. \(Y \sim \text{Binomial}(12, 0.19)\).

We need \(P(Y \geq 1) = 1 - P(Y = 0)\).

\(P(Y = 0) = (0.81)^{12}\)

Thus, \(P(Y \geq 1) = 1 - (0.81)^{12} = 0.920\).

(i) To find the value of c, we use the standard normal distribution. Given that 8% of carrots are shorter than c, we find the corresponding z-score for 0.08 in the standard normal distribution table, which is approximately -1.406.

Using the formula for standardization:

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

\(-1.406 = \frac{c - 14.2}{3.6}\)

Solving for c gives:

\(c = 14.2 + (-1.406 \times 3.6)\)

\(c = 9.14 \text{ cm}\)

(ii) To find the probability that at least 2 of the 7 carrots have lengths between 15 and 16 cm, we first find the probability that a single carrot is in this range.

Standardizing the values 15 and 16:

\(z_1 = \frac{15 - 14.2}{3.6} = 0.222\)

\(z_2 = \frac{16 - 14.2}{3.6} = 0.5\)

Using the standard normal distribution table, find:

\(P(0.222 < z < 0.5) = \Phi(0.5) - \Phi(0.222)\)

\(= 0.6915 - 0.5879 = 0.1036\)

Now, use the binomial distribution for 7 trials with probability 0.1036:

\(P(\text{at least 2}) = 1 - P(0) - P(1)\)

\(= 1 - (0.8964)^7 - 7 \times (0.8964)^6 \times 0.1036\)

\(= 1 - 0.8413\)

\(= 0.159\)

(a) To find the probability that the SBP of a randomly chosen adult is less than 132, we use the standard normal distribution. The z-score is calculated as:

\(z = \frac{132 - 125.4}{18.6} = 0.3548\)

The probability \(P(Z < 0.3548)\) is approximately 0.639.

(b) For the SBP of 12-year-old children, we know that 88% have SBP more than 108. This corresponds to a z-score of -1.175 (since 12% have SBP less than 108). Using the formula:

\(\frac{108 - 117}{\sigma} = -1.175\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{108 - 117}{-1.175} = 7.66\)

(c) To find the probability that each of these three adults has SBP within 1.5 standard deviations of the mean, we calculate:

\(P(-1.5 < Z < 1.5) = \Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1\)

Using \(\Phi(1.5) \approx 0.9332\), we have:

\(2 \times 0.9332 - 1 = 0.8664\)

The probability that each of the three adults has SBP within this range is:

\(0.8664^3 = 0.650\)

(i) To find the probability that a building is classified as tall, we need to calculate the probability that the height is greater than 70 metres. Using the standard normal distribution, we standardize the height:

\(z = \frac{70 - 50}{16} = 1.25\)

The probability \(P(z > 1.25)\) is found using the standard normal distribution table:

\(P(z > 1.25) = 1 - P(z \leq 1.25) = 1 - 0.8944 = 0.106\)

(ii) Let \(P(\text{short})\) be the probability that a building is classified as short. Since there are twice as many medium buildings as short ones, \(P(\text{medium}) = 2 \times P(\text{short})\). The total probability for medium and short buildings is:

\(P(\text{medium}) + P(\text{short}) = 1 - P(\text{tall}) = 1 - 0.106 = 0.894\)

Substituting \(P(\text{medium}) = 2 \times P(\text{short})\), we have:

\(2P(\text{short}) + P(\text{short}) = 0.894\)

\(3P(\text{short}) = 0.894\)

\(P(\text{short}) = \frac{0.894}{3} = 0.298\)

To find the height below which buildings are classified as short, we find the z-score corresponding to \(P(z < z_0) = 0.298\). From the standard normal distribution table, \(z_0 \approx -0.53\).

Standardizing, we have:

\(-0.53 = \frac{x - 50}{16}\)

Solving for \(x\):

\(x = 50 + (-0.53 \times 16) = 41.5\) metres