Question

A recent study reported that 29​% of the residents of a particular community lived in poverty. Suppose a random sample of 200 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.)

Answer 1

Answer:

a) The probability of having a sample of 200 people and having more than 33% of residents living in poverty is P=0.10312 or 10.3%.

b) The probability of having a sample of 200 people and having more than 24% of residents living in poverty is P=0.94091 or 94.1%.

Step-by-step explanation:

a) As the proportion for the population is μ=0.29 , we expect the porportion for the sample to be around p=0.29, with 50% of chances of being higher and 50% of chances of being lower than 0.29.

The probability of having a sample with 33% or more is less than 50%, because 0.33 is greater than the population proportion of 0.29.

To calculate P(p>0.33) you have to substract from the probability of having more than 29% poverty, which is 50%, the probability of having between 29% and 33%, which is not null.

We can write this as:

$P(p>0.33)=P(p>0.29)-P(0.29$

And we know $P(0.29$ is greater than zero, so $P(p>0.33)[\tex] is less than 0.5 or 50%.If we want to calculate the proability of having a p>0.33 with a sample of n=200, and assuming we can apply the central limit theorem (CLT), we have to calculate first the z-value:[tex]z=\frac{p-\bar{p}}{\sigma} =\frac{p-\bar{p}}{\sqrt{\frac{\bar{p}*(1-\bar{p})}{n}}} \\\\z=\frac{0.33-0.29}{\sqrt{\frac{0.29*(1-0.29)}{200}}}=\frac{0.04}{0.032}=1.246$

With this value of z, we can look up the probability of having p>0.33

$P(p>0.33)=P(z>1.246)=0.10312$

b) The probability of having a sample with 24% or more is more than 50%, because 0.24 is smaller than the population proportion of 0.29, so we add the 50% chances of being above the mean to the proability of having between 0.24 and 0.29 of poverty proportion, which is greater than zero.

If we want to calculate the proability of having a p>0.24 with a sample of n=200, and assuming we can apply the central limit theorem (CLT), we have to calculate first the z-value:

$z=\frac{p-\bar{p}}{\sigma} =\frac{p-\bar{p}}{\sqrt{\frac{\bar{p}*(1-\bar{p})}{n}}} \\\\z=\frac{0.24-0.29}{\sqrt{\frac{0.29*(1-0.29)}{200}}}=\frac{-0.05}{0.032}=-1.5625$

With this value of z, we can look up the probability of having p>0.24

$P(p>0.24)=P(z>-1.5625)=0.94091$