Respuesta :
Answer:
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%.
Hope this Helps!!!
The answer to whether the probability is greater than 50% or less than 50% is; less than 50% because 0.33 is greater than the population proportion
What is the probability of the normal distribution?
A) We want to determine the probability that 33% or more of our sample will be living in poverty and this is expressed as P(x ≥ 0.33)
Population proportion is; p = 0.28.
Formula for z-score is;
z = (x' - μ)/σ
Since our sample proportion is greater than our population proportion, it means the z-score will be greater than 0
Finally, due to the fact that 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 will definitely be less than 50%.
b) We want to now calculate the above probability;
P(x ≥ 0.33)
The sample mean will be equal to population mean as;
μₓ = μ = 0.28
Standard deviation of the distribution of sample means is;
σₓ = √(p(1 - p)/n)
σₓ = √((0.28×0.72)/300)
σₓ = 0.0259
Thus, z-score is;
z = (x - μ)/σ
z = (0.33 - 0.28)/0.0259
z = 1.93
Thus, the of 33% or more of the sample size is living in poverty is;
P(x ≥ 0.33) = P(z ≥ 1.93) = 1 - P(z < 1.93)
P(x ≥ 0.33) = 1 - 0.9732
P(x ≥ 0.33) = 0.0268
P(x ≥ 0.33) = 0.027
It is indeed way less than 50%.
Read more about normal distribution at; https://brainly.com/question/4079902
