The problem involves using the properties of the normal distribution. We're given that the lifetimes of light bulbs are normally distributed with a mean ($\mu$) of 56 hours and a standard deviation ($\sigma$) of 3.2 hours. The standard normal distribution (denoted by $Z$) is then used to standardize the values.

The formula for standardizing a variable $x$ is:

$Z = \frac{x - \mu}{\sigma}$

Let's calculate the required proportions step by step:

(a) What proportion of light bulbs will last more than 62 hours?

1. Standardize $x = 62$:

$Z = \frac{62 - 56}{3.2} = \frac{6}{3.2} = 1.875$

2. Find the corresponding $Z$-value from a standard normal distribution table or use a calculator:

   • The cumulative probability for $Z = 1.875$ is approximately 0.9696.

3. The proportion of light bulbs that last more than 62 hours is:

$P(X > 62) = 1 - P(Z \leq 1.875) = 1 - 0.9696 = 0.0304$

Thus, about 3.04% of light bulbs will last more than 62 hours.

(b) What proportion of light bulbs will last 53 hours or less?

1. Standardize $x = 53$:

$Z = \frac{53 - 56}{3.2} = \frac{-3}{3.2} = -0.9375$

2. From the standard normal distribution table, the cumulative probability for $Z = -0.9375$ is approximately 0.1748.

So, about 17.48% of light bulbs will last 53 hours or less.

(c) What proportion of light bulbs will last between 58 and 61 hours?

1. Standardize $x = 58$ and $x = 61$:

$Z_{58} = \frac{58 - 56}{3.2} = \frac{2}{3.2} = 0.625$ $Z_{61} = \frac{61 - 56}{3.2} = \frac{5}{3.2} = 1.5625$

2. Find the cumulative probabilities:

   • For $Z = 0.625$, the cumulative probability is approximately 0.734.
   • For $Z = 1.5625$, the cumulative probability is approximately 0.941.

3. The proportion of light bulbs lasting between 58 and 61 hours is:

$P(58 < X < 61) = P(Z \leq 1.5625) - P(Z \leq 0.625) = 0.941 - 0.734 = 0.207$

Thus, about 20.7% of light bulbs will last between 58 and 61 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 46 hours?

1. Standardize $x = 46$:

$Z = \frac{46 - 56}{3.2} = \frac{-10}{3.2} = -3.125$

2. The cumulative probability for $Z = -3.125$ is very small, approximately 0.0009.

Thus, the probability that a randomly selected light bulb lasts less than 46 hours is about 0.09%.

---

Would you like more details on any part of this? Here are some follow-up questions to consider:

1. How would the proportion change if the mean was higher than 56 hours?
2. What proportion of light bulbs will last between 50 and 60 hours?
3. How would we calculate the proportion if the standard deviation was doubled?
4. What does it mean if a light bulb lasts more than 70 hours in this distribution?
5. How would the distribution be affected if the light bulbs were manufactured with more precision?

Tip: Normal distribution tables and software tools like Excel or a calculator can help quickly find probabilities for different $Z$-values.