A recent survey found that 70% of all adults over 50 wear glasses for driving. You randomly select 30 adults over 50, and ask if he or she wears glasses. Decide whether you can use the normal distribution to approximate the binomial distribution. If so, find the mean and standard deviation. If not, explain why.

To determine if we can use the normal distribution to approximate the binomial distribution, we need to check for two conditions:

1. The number of trials (n) must be large, and
2. The probability of success (p) must not be too close to 0 or 1.

In this case, the number of trials (n = 30) is not exceptionally large, but we can still check the second condition.

The probability of success is given as 70%, which can be expressed as 0.7. Since this probability is not extremely close to 0 or 1, we can consider it acceptable to approximate the binomial distribution with the normal distribution.

To approximate the binomial distribution with the normal distribution, we can use the following formulas:

Mean (µ) = n * p
Standard deviation (σ) = sqrt(n * p * (1 - p))

Using the given values, we have:
Mean (µ) = 30 * 0.7 = 21
Standard deviation (σ) = sqrt(30 * 0.7 * (1 - 0.7)) = sqrt(6.3) ≈ 2.51

Therefore, we can approximate the binomial distribution with the normal distribution, with a mean of 21 and a standard deviation of approximately 2.51.

To determine whether you can use the normal distribution to approximate the binomial distribution in this scenario, there are two main criteria to consider:

1. The sample size should be large: The general rule is that for a binomial distribution to be approximated by a normal distribution, both the number of successes (individuals wearing glasses) and failures (individuals not wearing glasses) must be at least 5.

2. The probability of success should be neither extremely close to 0 nor 1: In this case, the probability of success is 70%, or 0.70.

Let's check if these criteria are met:

1. Sample size: You randomly select 30 adults over 50. As long as both the number of successes (wearing glasses) and failures (not wearing glasses) are at least 5, the sample size is considered large enough.

To check, let's calculate the expected number of successes and failures:

Number of successes (expected): 30 * 0.70 = 21

Number of failures (expected): 30 - 21 = 9

Since both the number of successes (21) and failures (9) are at least 5, the sample size of 30 is large enough to meet this criterion.

2. Probability of success: The probability of success (p) in this case is 0.70, which is not extremely close to either 0 or 1.

Given that both criteria are satisfied, you can use the normal distribution as an approximation for the binomial distribution in this case.

To find the mean (μ) and standard deviation (σ) for the normal distribution approximation, you can use the formulas:

Mean (μ) = n * p
Standard deviation (σ) = √(n * p * (1 - p))

where n is the sample size and p is the probability of success.

Substituting the values:

Mean (μ) = 30 * 0.70 = 21
Standard deviation (σ) = √(30 * 0.70 * (1 - 0.70)) = √(30 * 0.70 * 0.30) = √6.3 ≈ 2.51

Therefore, the mean (μ) for the normal distribution approximation is 21 and the standard deviation (σ) is approximately 2.51.

Mean = np = 30 * .7 = ?

Standard deviation = √npq = √(30 * .7 * .3) = ?

Note: q = 1 - p

I'll let you finish the calculations.