(i) To find the standard deviation \(\sigma\), use the standard normal distribution formula:

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

Given \(X = 4.2\), \(\mu = 3.9\), and \(P(X > 4.2) = 0.18\), find \(z\) such that \(P(Z > z) = 0.18\). From standard normal tables, \(z \approx 0.916\).

\(0.916 = \frac{4.2 - 3.9}{\sigma}\)

\(\sigma = \frac{0.3}{0.916} \approx 0.328\)

(ii) To find the probability that the length differs from the mean by less than half a minute, calculate:

\(z = \frac{4.4 - 3.9}{0.328} \approx 1.5267\)

\(z = \frac{3.4 - 3.9}{0.328} \approx -1.5267\)

Using standard normal tables, \(\Phi(1.5267) \approx 0.9364\).

Probability = \(2 \times 0.9364 - 1 = 0.873\)

(iii) Since the standard deviation is doubled, the spread is larger, resulting in a smaller \(z\)-value for the same difference from the mean. Therefore, \(p\) is less than the probability in part (ii).

(i) To find μ, we use the standard normal distribution. Given P(X < 6.8) = 0.75, we find the corresponding z-value from the standard normal distribution table, which is approximately z = 0.674.

Using the standardization formula:

\(z = \frac{6.8 - μ}{0.25μ}\)

\(Substitute z = 0.674:\)

\(0.674 = \frac{6.8 - μ}{0.25μ}\)

Solving for μ:

\(0.674 imes 0.25μ = 6.8 - μ\)

\(0.1685μ = 6.8 - μ\)

\(1.1685μ = 6.8\)

\(μ = \frac{6.8}{1.1685} \approx 5.82\)

(ii) To find P(X < 4.7), we standardize using the mean and standard deviation:

\(P(X < 4.7) = P\left( z < \frac{4.7 - 5.82}{1.4548} \right)\)

\(= P(z < -0.769)\)

Using the standard normal distribution table, \(P(z < -0.769) = 1 - 0.7791 = 0.221\)

Given that 25% of women have fingers longer than 8.8 cm, the corresponding z-score is \(z = 0.674\) (since 75% is below 8.8 cm).

Using the z-score formula: \(z = \frac{X - \mu}{\sigma}\), we have:

\(0.674 = \frac{8.8 - \mu}{\sigma}\)

\(0.674\sigma = 8.8 - \mu\)

Similarly, for 17.5% having fingers shorter than 7.7 cm, the z-score is \(z = -0.935\).

\(-0.935 = \frac{7.7 - \mu}{\sigma}\)

\(-0.935\sigma = 7.7 - \mu\)

We now have two equations:

1. \(0.674\sigma = 8.8 - \mu\)
2. \(-0.935\sigma = 7.7 - \mu\)

Subtract the second equation from the first:

\(0.674\sigma + 0.935\sigma = 8.8 - 7.7\)

\(1.609\sigma = 1.1\)

\(\sigma = \frac{1.1}{1.609} \approx 0.684\)

Substitute \(\sigma = 0.684\) back into the first equation:

\(0.674 \times 0.684 = 8.8 - \mu\)

\(0.461 = 8.8 - \mu\)

\(\mu = 8.8 - 0.461\)

\(\mu = 8.34\)

(i) To find the proportion of bananas that are small, we standardize the weight 95 grams using the formula:

\(z = \frac{95 - 150}{50} = -1.1\)

We find \(P(z < -1.1)\) using the standard normal distribution table:

\(P(z < -1.1) = 0.136\)

Thus, the proportion of bananas that are small is 0.136.

(ii) To find the weight exceeded by 10% of bananas, we need the 90th percentile of the distribution. The z-value for 90% is approximately 1.282.

Using the standardization formula:

\(1.282 = \frac{x - 150}{50}\)

Solving for \(x\):

\(x = 1.282 \times 50 + 150 = 214 \text{ g}\)

(iii) (a) The probability that a banana is medium is given by:

\(P(\text{medium}) = 1 - 2 \times 0.1357 = 0.7286\)

(b) The expected cost per banana is calculated as follows:

\(0.1357 \times 10 + 0.1357 \times 25 + 0.7286 \times 20 = 19.3215 \text{ cents}\)

The total cost for 100 bananas is:

\(100 \times 19.3215 = 1930 \text{ cents} = \$19.30\)

Given that the time to cook an egg is normally distributed with mean \(\mu = 4.2\) minutes and standard deviation \(\sigma = 0.6\) minutes. We need to find the value of \(t\) such that 12% of people take more than \(t\) minutes.

This implies that 88% of people take less than \(t\) minutes. We find the corresponding \(z\)-score for 0.88 in the standard normal distribution table, which is approximately \(z = 1.175\).

Using the standardization formula:

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

Substitute the known values:

\(1.175 = \frac{t - 4.2}{0.6}\)

Solving for \(t\):

\(t - 4.2 = 1.175 \times 0.6\)

\(t - 4.2 = 0.705\)

\(t = 4.2 + 0.705\)

\(t = 4.905\)

Rounding to two decimal places, \(t = 4.91\).