A recent study reported that 28% of the residents of a particular community lived in poverty. Suppose a random sample of 300 residents of this community is taken. We wish to determine the probability that 33% or more of our sample will be living in poverty. Complete parts (a) and (b) below.

Before doing any calculations, determine whether this probability is greater than 50% or less than 50%. Why? The answer should be less than 50%, because the resulting z-score will be negative and the sampling distribution is approximately Normal. The answer should be greater than 50%, because 0.24 is greater than the population proportion of 0.20 and because the sampling distribution is approximately Normal. The answer should be less than 50%, because 0.24 is greater than the population proportion of 0.20 and because the sampling distribution is approximately Normal. The answer should be greater than 50%, because the resulting z-score will be positive and the sampling distribution is approximately Normal. Calculate the probability that 24% or more of the sample will be living in poverty. Assume the sample is collected in such a way that the conditions for using the CLT are met. P (p ge 0.24) = (Round to three decimal places as needed.)

a) The answer should be less than 50%, because 0.33 is greater than the population proportion of 0.28 and because the sampling distribution is approximately Normal.

b) P(x ≥ 0.33) = 2.68% = 0.027 to 3 d.p

Step-by-step explanation:

a) The required probability is P(x ≥ 0.33)

The population proportion of people living in poverty has already been given as 0.28, So, whatever the standard deviation of the distribution of sample means is, the sample proportion of 0.33 is more than the population proportion, So, it gives a z-score that is greater than 0. The probability from the z-score of the mean (0) to the end of distribution is exactly 50%; So, a Probability that only covers from a particular positive z-score to the end of the distribution will definitely be less than 50%.

So, because 33% is more than population proportion of 28%, and the sampling distribution is approximately normal, the probability of 33% or more of the sample living in poverty is less than 50%.

b) P(x ≥ 0.33)

The sample mean = population mean

μₓ = μ = 0.28

Standard deviation of the distribution of sample means = √[p(1-p)/n] (this is possible due to the central limit theorem for n greater than 30)

σₓ = √[(0.28×0.72)/300] = 0.0259

We then normalize 0.33

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.33 - 0.28)/0.0259 = 1.93

To determine the probability of 33% or more of the sample size is living in poverty.

P(x ≥ 0.33) = P(z ≥ 1.93)

We'll use data from the normal probability table for these probabilities

P(x ≥ 0.33) = P(z ≥ 1.93) = 1 - P(z < 1.93)

= 1 - 0.9732 = 0.0268 = 2.68%

It is indeed way less than 50%.

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step-by-step explanation: