Respuesta :
Answer:
86.64% probability that the resulting sample proportion is within .02 of the true proportion.
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 p in a sample of size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem:
[tex]\mu = 0.2, \sigma = \sqrt{\frac{0.2*0.8}{900}} = 0.0133[/tex]
How likely is the resulting sample proportion to be within .02 of the true proportion (i.e., between .18 and .22)?
This is the pvalue of Z when X = 0.22 subtracted by the pvalue of Z when X = 0.18.
X = 0.22
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.22 - 0.2}{0.0133}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332.
X = 0.18
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.18 - 0.2}{0.0133}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
0.9332 - 0.0668 = 0.8664
86.64% probability that the resulting sample proportion is within .02 of the true proportion.
