In a local university, 40% of the students live in the dormitories. A random sample of 80 students is selected for a particular study. The probability that the sample proportion (the proportion living in the dormitories) is at least 0.30 is

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.9664

To find the probability that the sample proportion is at least 0.30, we can use the normal distribution approximation to the binomial distribution.

First, let's calculate the mean and standard deviation of the sample proportion.

Given that 40% of the students live in the dormitories, the mean of the sample proportion is equal to the population proportion, which is 0.40.

The standard deviation of the sample proportion can be calculated using the formula:
σ = sqrt((p * (1 - p)) / n)
where p is the population proportion (0.40) and n is the sample size (80).

σ = sqrt((0.40 * (1 - 0.40)) / 80)
= sqrt((0.24) / 80)
≈ 0.0548

Now, we can convert the problem into a normal distribution by using the z-score.

The z-score for a given sample proportion value can be calculated using the formula:
z = (x - μ) / σ
where x is the sample proportion value (0.30), μ is the mean of the sample proportion (0.40), and σ is the standard deviation of the sample proportion (0.0548).

z = (0.30 - 0.40) / 0.0548
≈ -1.825

To find the probability that the sample proportion is at least 0.30, we need to find the area under the normal distribution curve to the left of the given z-score and subtract it from 1.

P(z ≥ -1.825) = 1 - P(z < -1.825)

Using a standard normal distribution table or a calculator, we can find that P(z < -1.825) ≈ 0.0336.

Therefore, the probability that the sample proportion is at least 0.30 is approximately:
P(z ≥ -1.825) = 1 - P(z < -1.825) ≈ 1 - 0.0336 = 0.9664.

So, the probability that the sample proportion is at least 0.30 is approximately 0.9664 or 96.64%.