We are given that 35% of customers shop online, so the probability of a customer shopping online is 0.35. For a sample of 100 customers, we can model the number of customers shopping online as a binomial distribution with parameters \(n = 100\) and \(p = 0.35\).

To approximate this binomial distribution, we use a normal distribution with the same mean and variance. The mean \(\mu\) is given by:

\(\mu = np = 100 \times 0.35 = 35\)

The variance \(\sigma^2\) is given by:

\(\sigma^2 = np(1-p) = 100 \times 0.35 \times 0.65 = 22.75\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{22.75}\).

We need to find \(P(X > 39)\). Using the normal approximation, we apply a continuity correction and find \(P(X > 39.5)\).

We standardize this to find the corresponding \(z\)-score:

\(z = \frac{39.5 - 35}{\sqrt{22.75}} = 0.943\)

We then find \(P(Z > 0.943)\) using the standard normal distribution table:

\(P(Z > 0.943) = 1 - P(Z < 0.943) = 1 - 0.8272 = 0.173\)

Therefore, the probability that more than 39 customers shop online is 0.173.

(ii) The probability that a light bulb has a lifetime of more than 2500 hours is 0.2. For 300 bulbs, the mean is calculated as:

\(\text{Mean} = 300 \times 0.2 = 60\)

The variance is:

\(\text{Variance} = 300 \times 0.2 \times 0.8 = 48\)

We use the normal approximation to the binomial distribution. The probability that fewer than 70 bulbs have a lifetime of more than 2500 hours is:

\(P(X < 70) = P\left(Z > \frac{69.5 - 60}{\sqrt{48}}\right)\)

\(= \Phi(1.371)\)

\(= 0.915\)

(iii) The conditions for using the normal approximation are met because:

\(np = 60\) and \(nq = 240\), both greater than 5.

Let the random variable \(X\) represent the number of students who pass the examination. \(X\) follows a binomial distribution with parameters \(n = 200\) and \(p = 0.8\).

To approximate this binomial distribution, we use a normal distribution because both \(np\) and \(n(1-p)\) are greater than 5.

The mean of the binomial distribution is \(np = 200 \times 0.8 = 160\).

The variance is \(np(1-p) = 200 \times 0.8 \times 0.2 = 32\).

Thus, the standard deviation is \(\sqrt{32}\).

We approximate \(X\) by a normal distribution \(N(160, 32)\).

We need to find \(P(X > 166)\). Using continuity correction, this is approximately \(P(X > 166.5)\).

Standardizing, we have:

\(z = \frac{166.5 - 160}{\sqrt{32}} \approx 1.149\)

Thus, \(P(X > 166.5) = P(Z > 1.149) = 1 - P(Z \leq 1.149)\).

Using standard normal distribution tables, \(P(Z \leq 1.149) \approx 0.8747\).

Therefore, \(P(Z > 1.149) = 1 - 0.8747 = 0.125\).

For part (iii), the justification for using the normal approximation is that both \(np = 160\) and \(nq = 40\) are greater than 5, which satisfies the conditions for the normal approximation to the binomial distribution.

(i) To find the probability that an apple can be used as a toffee apple, we calculate the probability that the diameter is between 4.1 cm and 5 cm. We standardize these values using the formula for the z-score:

\(z_1 = \frac{4.1 - 5.7}{0.8} = -2\)

\(z_2 = \frac{5 - 5.7}{0.8} = -0.875\)

The probability is given by:

\(P(4.1 < d < 5) = P(z < -0.875) - P(z < -2)\)

Using standard normal distribution tables:

\(\Phi(-0.875) = 0.1908\)

\(\Phi(-2) = 0.0228\)

Thus, \(P(4.1 < d < 5) = 0.1908 - 0.0228 = 0.168\)

(ii) For 250 apples, the number that can be used as toffee apples follows a binomial distribution with \(n = 250\) and \(p = 0.168\). We approximate this with a normal distribution:

\(np = 250 \times 0.168 = 42\)

\(npq = 34.944\)

Standard deviation \(\sigma = \sqrt{34.944} = 5.911\)

We find \(P(X < 50)\) using continuity correction:

\(P(X < 50) = P\left( z < \frac{49.5 - 42}{5.911} \right) = P(z < 1.2687)\)

Using standard normal distribution tables:

\(\Phi(1.2687) = 0.898\)

Let the random variable \(X\) represent the number of people who choose a mobile phone made by Company B. \(X\) follows a binomial distribution with parameters \(n = 130\) and \(p = 0.35\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 130 \times 0.35 = 45.5\)

\(\sigma^2 = np(1-p) = 130 \times 0.35 \times 0.65 = 29.575\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{29.575} \approx 5.438\).

We need to find \(P(X \geq 50)\). Using the normal approximation, we standardize:

\(P(X \geq 50) = P\left( Z \geq \frac{49.5 - 45.5}{5.438} \right) = P(Z \geq 0.7355)\)

Using the standard normal distribution table, \(P(Z \geq 0.7355) = 1 - \Phi(0.7355) \approx 1 - 0.7691 = 0.231\).