HELP ME WITH THIS PLEASE
2 answers:
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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, <em>with 50% of chances of being higher and 50% of chances of being lower than 0.29</em>.
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:
And we know is greater than zero, so
With this value of z, we can look up the probability of having p>0.33
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:
With this value of z, we can look up the probability of having p>0.24