To solve this problem, we need to calculate the mean and standard deviation of a binomial distribution. In this case, the binomial distribution corresponds to the number of adults (out of 100) who believe the state of moral values is poor.

We are given the following information:

• $n = 100$ (the number of adults surveyed)
• $p = 0.54$ (the probability that an adult believes the state of moral values is poor)
• $q = 1 - p = 0.46$ (the complement of the probability)

Part (a): Mean of $X$

For a binomial distribution, the mean ($\mu$) is given by:

$\mu = n \times p$

Substituting the values:

$\mu = 100 \times 0.54 = 54$

So, the mean number of adults who believe the overall state of moral values is poor is 54.

Part (b): Standard deviation of $X$

The standard deviation ($\sigma$) of a binomial distribution is calculated using the formula:

$\sigma = \sqrt{n \times p \times q}$

Substituting the values:

$\sigma = \sqrt{100 \times 0.54 \times 0.46} \approx \sqrt{24.84} \approx 4.98$

Rounding to the nearest tenth, the standard deviation is 5.0.

Final Answers:

• Mean = 54
• Standard deviation = 5.0

Would you like further clarification or details on how the formulas were applied? Here are some related questions for further practice:

1. How would the mean and standard deviation change if the survey was conducted with a sample of 200 adults instead of 100?
2. What happens to the mean and standard deviation if the percentage of adults believing moral values are poor changes to 60%?
3. How do you compute the probability of observing exactly 60 adults out of 100 who believe that moral values are poor using the binomial distribution?
4. How do the formulas for mean and standard deviation change if the survey asks about a different characteristic, such as political affiliation?
5. Can you explain the concept of variance and how it relates to standard deviation?

Tip: In binomial distributions, the mean gives you the expected number of successes, and the standard deviation helps you measure how much the actual number of successes might vary from this expected value.