Respuesta :
Answer:
0.9146 = 91.46% probability that the proportion of persons with a college degree will differ from the population proportion by less than 4%
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
For the sampling distribution of a sample proportion of size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem, we have that:
[tex]\mu = 0.47, \sigma = \sqrt{\frac{0.47*0.53}{460}} = 0.0233[/tex]
What is the probability that the proportion of persons with a college degree will differ from the population proportion by less than 4%?
Population proportion between 0.47-0.04 = 0.43 and 0.47+0.04 = 0.51, which is the pvalue of Z when X = 0.51 subtracted by the pvalue of Z when X = 0.43. So
X = 0.51
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.51 - 0.47}{0.0233}[/tex]
[tex]Z = 1.72[/tex]
[tex]Z = 1.72[/tex] has a pvalue of 0.9573
X = 0.43
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.43 - 0.47}{0.0233}[/tex]
[tex]Z = -1.72[/tex]
[tex]Z = -1.72[/tex] has a pvalue of 0.0427
0.9573 - 0.0427 = 0.9146
0.9146 = 91.46% probability that the proportion of persons with a college degree will differ from the population proportion by less than 4%
