(ii) To approximate the probability, we use the normal approximation to the binomial distribution. Given that the sample size is 500, we calculate:

Mean: \(np = 340\)

Variance: \(npq = 108.8\)

We standardize the variable using continuity correction:

\(P(x > 337) = P\left( z > \frac{337.5 - 340}{\sqrt{108.8}} \right)\)

\(= P(z > -0.2396)\)

Using the standard normal distribution table, we find:

\(P(z > -0.2396) = 0.595\)

(iii) The normal approximation is justified because both \(np = 340\) and \(nq = 160\) are greater than 5.

Let the number of adults who own a cell phone be a binomial random variable with parameters \(n = 160\) and \(p = 0.75\).

Calculate the mean and variance:

\(np = 160 \times 0.75 = 120\)

\(npq = 160 \times 0.75 \times 0.25 = 30\)

Use the normal approximation to the binomial distribution:

\(P(X > 114) = P\left( z > \frac{114.5 - 120}{\sqrt{30}} \right)\)

\(= P(z > -1.004)\)

\(= \Phi(1.004) = 0.842\)

Thus, \(P(X > 114) = 1 - 0.842 = 0.158\).

For part (iii), the approximation is justified because both \(np\) and \(nq\) are greater than 5.

Let \(X\) be the number of days Wenjie goes out with her friends. \(X\) follows a binomial distribution with \(n = 252\) and \(p = \frac{1}{7}\).

For normal approximation, calculate the mean \(\mu\) and variance \(\sigma^2\):

\(\mu = np = 252 \times \frac{1}{7} = 36\)

\(\sigma^2 = npq = 252 \times \frac{1}{7} \times \frac{6}{7} = 30.857\)

Standard deviation \(\sigma = \sqrt{30.857}\).

Using continuity correction, find:

\(P(X < 30) = P\left( z < \frac{29.5 - 36}{\sqrt{30.857}} \right)\)

\(P(X > 44) = P\left( z > \frac{44.5 - 36}{\sqrt{30.857}} \right)\)

Calculate the z-scores:

\(z_1 = \frac{29.5 - 36}{\sqrt{30.857}} = -1.170\)

\(z_2 = \frac{44.5 - 36}{\sqrt{30.857}} = 1.530\)

Using standard normal distribution tables:

\(P(z < -1.170) = 0.1209\)

\(P(z > 1.530) = 0.0630\)

Total probability: \(P(X < 30) + P(X > 44) = 0.1209 + 0.0630 = 0.184\)

For part (ii), the use of a normal approximation is justified because both \(np\) and \(nq\) are greater than 5.

(i) Given that 80% of the days, Rafa spends more than 1.35 hours, we have:

\(P(X > 1.35) = 0.8\)

Using the standard normal distribution, \(z = -0.842\) for 20% in the left tail:

\(P\left( \frac{1.35 - 1.9}{\sigma} < z \right) = 0.2\)

\(-0.842 = \frac{1.35 - 1.9}{\sigma}\)

\(\sigma = \frac{-0.55}{-0.842} = 0.653\)

(ii) To find \(P(X < 2)\):

\(P\left( z < \frac{2 - 1.9}{0.6532} \right) = P(z < 0.1531)\)

Using the standard normal table, \(P(z < 0.1531) = 0.561\).

(iii) For a sample of 200 days, \(X \sim N(160, 32)\):

We need \(P(162.5 < X < 173.5)\):

\(P\left( \frac{162.5 - 160}{\sqrt{32}} < z < \frac{173.5 - 160}{\sqrt{32}} \right)\)

\(P(0.442 < z < 2.386) = \Phi(2.386) - \Phi(0.442)\)

\(= 0.9915 - 0.6707 = 0.321\)

The probability that a person is neither a student, office worker, nor shop assistant is given by:

\(1 - (0.36 + 0.22 + 0.29) = 0.13\)

For a binomial distribution with parameters \(n = 300\) and \(p = 0.13\), the mean and variance are:

\(\text{mean} = 300 \times 0.13 = 39\)

\(\text{variance} = 300 \times 0.13 \times 0.87 = 33.93\)

We approximate the binomial distribution with a normal distribution \(N(39, 33.93)\).

Using continuity correction, we find:

\(P(31 \, \text{to} \, 49) = P(30.5 < x < 49.5)\)

Standardizing, we have:

\(P\left( \frac{30.5 - 39}{\sqrt{33.93}} < z < \frac{49.5 - 39}{\sqrt{33.93}} \right)\)

\(= P(-1.4592 < z < 1.8026)\)

Using the standard normal distribution table:

\(= \Phi(1.8026) + \Phi(1.4592) - 1\)

\(= 0.9643 + 0.9278 - 1\)

\(= 0.892\)