Let the random variable \(X\) represent the number of families owning a dishwasher in a sample of 145 families. \(X\) follows a binomial distribution with parameters \(n = 145\) and \(p = 0.22\).

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

Calculate the mean \(\mu\) and variance \(\sigma^2\):

\(\mu = 145 \times 0.22 = 31.9\)

\(\sigma^2 = 145 \times 0.22 \times 0.78 = 24.882\)

We use a continuity correction for \(P(X > 26)\), which becomes \(P(X > 26.5)\).

Standardize the variable:

\(P(X > 26.5) = P\left( z > \frac{26.5 - 31.9}{\sqrt{24.882}} \right) = P(z > -1.08255)\)

Using the standard normal distribution table, \(\Phi(1.08255) = 0.861\).

Thus, the probability that more than 26 families own a dishwasher is \(0.861\).

Let \(p = 0.25\) and \(n = 160\). The mean of the distribution is \(n \times p = 160 \times 0.25 = 40\).

The variance is \(n \times p \times (1-p) = 160 \times 0.25 \times 0.75 = 30\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(40, 30)\).

To find \(P(X < 50)\), we use the continuity correction: \(P(X < 49.5)\).

Standardize: \(Z = \frac{49.5 - 40}{\sqrt{30}} \approx 1.734\).

Thus, \(P(Z < 1.734) = 0.959\).

(a) To find the percentage of eggs classified as small, we calculate the probability that an egg weighs less than 76 grams. Using the standardization formula:

\(P(X < 76) = P\left(Z < \frac{76 - 80.5}{6.6}\right)\)

\(= P(Z < -0.6818) = 1 - \Phi(0.6818)\)

\(= 1 - 0.7524 = 0.2476\)

Thus, the percentage is 24.8%.

(b) To find the least possible weight of a large egg, we first find the percentage of large eggs:

\(\text{Percentage of large eggs} = 100 - 40 - 24.76 = 35.24\)

We then find the weight corresponding to the cumulative probability of 0.6476:

\(P(Z > x) = 0.6476\)

\(\frac{x - 80.5}{6.6} = 0.378\)

\(x = 80.5 + 0.378 \times 6.6 = 83.0\)

(c) For 150 eggs, the mean number classified as medium is:

\(\text{Mean} = 150 \times 0.4 = 60\)

\(\text{Variance} = 150 \times 0.4 \times 0.6 = 36\)

We approximate the probability that more than 68 eggs are medium:

\(P(X > 68) = P\left(Z > \frac{68.5 - 60}{\sqrt{36}}\right)\)

\(= P(Z > 1.417) = 1 - \Phi(1.417)\)

\(= 1 - 0.9217 = 0.0783\)

(i) The number of faulty CDs in a box follows a binomial distribution with parameters \(n = 30\) and \(p = 0.04\). We need to find \(P(X > 2)\), where \(X\) is the number of faulty CDs.

\(P(X > 2) = 1 - P(X \leq 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))\)

\(P(X = 0) = \binom{30}{0} (0.04)^0 (0.96)^{30}\)

\(P(X = 1) = \binom{30}{1} (0.04)^1 (0.96)^{29}\)

\(P(X = 2) = \binom{30}{2} (0.04)^2 (0.96)^{28}\)

Calculating these probabilities:

\(P(X = 0) = 0.2938\)

\(P(X = 1) = 0.3673\)

\(P(X = 2) = 0.2219\)

\(P(X > 2) = 1 - (0.2938 + 0.3673 + 0.2219) = 0.117\)

(ii) Let \(Y\) be the number of rejected boxes out of 280. \(Y\) follows a binomial distribution with parameters \(n = 280\) and \(p = 0.117\) (from part (i)).

We approximate \(Y\) using a normal distribution with mean \(\mu = np = 280 \times 0.117 = 32.73\) and variance \(\sigma^2 = np(1-p) = 280 \times 0.117 \times 0.883 = 28.9\).

Using the normal approximation, we find \(P(Y \geq 30)\) with continuity correction:

\(P(Y \geq 30) = P\left( Z \geq \frac{29.5 - 32.73}{\sqrt{28.9}} \right) = P(Z \geq -0.6008)\)

Using standard normal distribution tables, \(P(Z \geq -0.6008) = 0.726\).

Let \(X\) be the number of days George goes swimming in 270 days. \(X\) follows a binomial distribution with parameters \(n = 270\) and \(p = \frac{1}{3}\).

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

Calculate \(np\) and \(npq\):

\(np = 270 \times \frac{1}{3} = 90\)

\(npq = 270 \times \frac{1}{3} \times \frac{2}{3} = 60\)

We use a continuity correction to approximate \(P(X > 100)\) as \(P(X > 99.5)\).

Standardize using the normal distribution:

\(P(X > 99.5) = P\left( z > \frac{99.5 - 90}{\sqrt{60}} \right) = P(z > 1.2264)\)

Using the standard normal distribution table, find \(P(z > 1.2264) = 1 - P(z < 1.2264) = 1 - 0.8899 = 0.110\).