Let \(X\) be the number of years out of 56 where New Year's Day is on a Saturday. \(X\) follows a binomial distribution with parameters \(n = 56\) and \(p = \frac{1}{7}\).

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

\(\mu = 56 \times \frac{1}{7} = 8\)

\(\sigma^2 = 56 \times \frac{1}{7} \times \frac{6}{7} = 6.857\)

We need to find \(P(X > 7)\). Using a continuity correction, this becomes \(P(X > 7.5)\).

Standardizing, we have:

\(P(X > 7.5) = 1 - \Phi \left( \frac{7.5 - 8}{\sqrt{6.857}} \right)\)

\(= 1 - \Phi(-0.1909)\)

\(= \Phi(0.1909)\)

Using standard normal distribution tables, \(\Phi(0.1909) \approx 0.576\).

Let the number of small bands be a random variable, denoted by \(X\). Given that 60% of the bands are small, \(X\) follows a binomial distribution \(B(n=150, p=0.6)\).

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

\(\mu = np = 150 \times 0.6 = 90\)

\(\sigma^2 = np(1-p) = 150 \times 0.6 \times 0.4 = 36\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{36} = 6\).

We use a normal approximation to find \(P(88 < X < 97)\). Applying continuity correction, this becomes \(P(87.5 < X < 97.5)\).

Standardizing, we calculate:

\(P(87.5 < X < 97.5) = \Phi\left(\frac{97.5 - 90}{6}\right) - \Phi\left(\frac{87.5 - 90}{6}\right)\)

\(= \Phi(1.25) - \Phi(-0.4166)\)

Using standard normal distribution tables, \(\Phi(1.25) = 0.8944\) and \(\Phi(-0.4166) = 1 - 0.6616 = 0.3384\).

Thus, the probability is:

\(0.8944 - 0.3384 = 0.556\)

Let the random variable \(X\) represent the number of adults wearing a watch on their left wrist. Given that 76% wear a watch on their left wrist, \(X\) follows a binomial distribution \(B(n=200, p=0.76)\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are calculated as follows:

\(\mu = np = 200 \times 0.76 = 152\)

\(\sigma^2 = np(1-p) = 200 \times 0.76 \times 0.24 = 36.48\)

We use a continuity correction to find \(P(X > 155)\), which is approximated by \(P(X > 155.5)\) in the normal distribution.

Standardizing, we have:

\(Z = \frac{155.5 - 152}{\sqrt{36.48}} = \frac{3.5}{6.036} \approx 0.5795\)

Thus, \(P(X > 155) = 1 - \Phi(0.5795) \approx 1 - 0.7188 = 0.281\).

(i) We know that 10% of the bulbs last longer than 5130 hours. This corresponds to a z-score of 1.282 in a standard normal distribution. Using the formula for the z-score:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 5130\), \(\sigma = 40.6\), and \(z = 1.282\), we solve for \(\mu\):

\(1.282 = \frac{5130 - \mu}{40.6}\)

\(\mu = 5130 - 1.282 \times 40.6\)

\(\mu = 5080\) (rounded to the nearest hour)

(ii) To find the probability that a light bulb fails to last for 5000 hours, we calculate the z-score:

\(z = \frac{5000 - 5078}{40.6} = -1.921\)

The probability \(P(X < 5000) = \Phi(-1.921) = 0.0273\) or 2.73%.

(iii) For 600 light bulbs, the mean number lasting longer than 5130 hours is \(\mu = 60\) and variance \(\sigma^2 = 54\). We approximate using a normal distribution:

\(P(\text{fewer than 65}) = \Phi\left(\frac{64.5 - 60}{\sqrt{54}}\right)\)

\(= \Phi(0.6123)\)

\(= 0.730\) or 0.73

Let the random variable \(X\) represent the number of overweight people in the sample of 400. Given that 2 out of 5 people are overweight, the probability \(p\) of a person being overweight is \(p = \frac{2}{5} = 0.4\).

The expected number of overweight people \(\mu\) is given by \(\mu = np = 400 \times 0.4 = 160\).

The variance \(\sigma^2\) is given by \(\sigma^2 = np(1-p) = 400 \times 0.4 \times 0.6 = 96\).

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is \(\sqrt{96} \approx 9.798\).

We apply continuity correction for the normal approximation: \(P(X < 165) \approx P(X \leq 164.5)\).

Standardizing, we find:

\(Z = \frac{164.5 - 160}{\sqrt{96}} = \frac{4.5}{9.798} \approx 0.4593\)

Using standard normal distribution tables, \(\Phi(0.4593) \approx 0.677\).

Thus, the probability that fewer than 165 people in the sample are overweight is approximately 0.677.