Respuesta :
Answer:
99.96% probability that the sample proportion will be within 10 percent of the population 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.
In this problem, we have that:
Proportion p = 0.75
Mean:
[tex]\mu = p = 0.75[/tex]
Standard deviation of the proportion:
[tex]\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.75*0.25}{229}} = 0.0286[/tex]
What is the probability that the sample proportion will be within 10 percent of the population proportion?
This is the pvalue of Z when X = 0.75+0.1 = 0.85 subtracted by the pvalue of Z when X = 0.75 - 0.1 = 0.65. So
X = 0.85
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.85 - 0.75}{0.0286}[/tex]
[tex]Z = 3.49[/tex]
[tex]Z = 3.49[/tex] has a pvalue of 0.9998
X = 0.65
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.65 - 0.75}{0.0286}[/tex]
[tex]Z = -3.49[/tex]
[tex]Z = -3.49[/tex] has a pvalue of 0.0002
0.9998 - 0.0002 = 0.9996
99.96% probability that the sample proportion will be within 10 percent of the population proportion
