(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:

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