Let the probability that a person does not support either choir be denoted by \(p\). Given that 30% support Notes and 45% support Classics, the probability of not supporting either is:

\(p = 1 - 0.30 - 0.45 = 0.25\)

The mean number of people in the sample who do not support either choir is:

\(\text{Mean} = 240 \times 0.25 = 60\)

The variance is:

\(\text{Variance} = 240 \times 0.25 \times 0.75 = 45\)

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is:

\(\sigma = \sqrt{45}\)

We apply the continuity correction for \(X < 50\):

\(P(X < 50) = P\left( Z < \frac{49.5 - 60}{\sqrt{45}} \right) = P(Z < -1.565)\)

Using standard normal distribution tables, \(P(Z < -1.565)\) is approximately:

\(1 - 0.9412 = 0.0588\)

Thus, the probability that fewer than 50 do not support either choir is \(0.0588\).

Let the random variable \(X\) represent the number of faulty components in a sample of 200. \(X\) follows a binomial distribution with parameters \(n = 200\) and \(p = 0.15\).

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

\(\mu = np = 200 \times 0.15 = 30\)

\(\sigma^2 = np(1-p) = 200 \times 0.15 \times 0.85 = 25.5\)

Thus, \(X\) is approximately \(N(30, 25.5)\).

We need to find \(P(X > 40)\). Using a continuity correction, this is equivalent to \(P(X > 40.5)\).

Standardizing, we have:

\(P(X > 40.5) = P\left(Z > \frac{40.5 - 30}{\sqrt{25.5}}\right)\)

\(= P\left(Z > \frac{10.5}{5.04975}\right)\)

\(= P(Z > 2.079)\)

Using standard normal distribution tables or a calculator, \(P(Z > 2.079) = 1 - \Phi(2.079) \approx 1 - 0.9812 = 0.0188\).

Let the random variable \(X\) represent the number of households satisfied with their wifi connection. \(X\) follows a binomial distribution with parameters \(n = 150\) and \(p = 0.66\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(np, npq)\), where \(q = 1 - p\).

Calculate the mean and variance:

\(np = 0.66 \times 150 = 99\)

\(npq = 0.66 \times (1 - 0.66) \times 150 = 33.66\)

Standard deviation \(\sigma = \sqrt{33.66} \approx 5.8017\)

We need to find \(P(X > 84)\). Using continuity correction, this is \(P(X > 84.5)\).

Standardize using the normal distribution:

\(P(X > 84.5) = P\left( Z > \frac{84.5 - 99}{\sqrt{33.66}} \right)\)

\(= P(Z > -2.499)\)

Using standard normal distribution tables or a calculator, \(P(Z > -2.499) = 0.994\).

Let the random variable representing the number of men be denoted by a binomial distribution, \(X \sim B(n, p)\), where \(n = 600\) and \(p = 0.34\).

The mean \(\mu\) is given by \(np = 600 \times 0.34 = 204\).

The variance \(\sigma^2\) is given by \(np(1-p) = 600 \times 0.34 \times 0.66 = 134.64\).

We approximate the binomial distribution with a normal distribution \(N(\mu, \sigma^2)\).

To find \(P(X < 190)\), we use a continuity correction: \(P(X < 190) \approx P(X < 189.5)\).

Standardize using \(Z = \frac{X - \mu}{\sigma}\):

\(P(X < 189.5) = P\left( Z < \frac{189.5 - 204}{\sqrt{134.64}} \right) = P(Z < -1.2496)\).

Using the standard normal distribution table, \(P(Z < -1.2496) = 1 - \Phi(1.2496)\).

\(\Phi(1.2496) \approx 0.8944\), so \(1 - 0.8944 = 0.106\).

The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 250\)

- Probability of success, \(p = 0.65\)

We approximate the binomial distribution with a normal distribution using the following parameters:

- Mean, \(\mu = np = 250 \times 0.65 = 162.5\)

- Variance, \(\sigma^2 = np(1-p) = 250 \times 0.65 \times 0.35 = 56.875\)

- Standard deviation, \(\sigma = \sqrt{56.875}\)

We need to find \(P(X < 179)\). Using the continuity correction, we approximate this as \(P(X < 178.5)\).

Standardize the variable:

\(Z = \frac{178.5 - 162.5}{\sqrt{56.875}} = \frac{16}{\sqrt{56.875}} \approx 2.122\)

Using the standard normal distribution table, find \(P(Z < 2.122)\).

The probability is approximately \(0.983\).