Respuesta :
Answer:
95.86% probability that the sample proportion will differ from the population proportion by less than 3%
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:
[tex]p = 0.07[/tex]
For a proportion, we have that:
[tex]\mu = p = 0.07[/tex]
[tex]\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.07*0.93}{300}} = 0.0147[/tex]
What is the probability that the sample proportion will differ from the population proportion by less than 3%
This is the pvalue of Z when X = 0.07 + 0.03 = 0.1 subtracted by the pvalue of Z when X = 0.07 - 0.03 = 0.04. So
X = 0.1
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.1 - 0.07}{0.0147}[/tex]
[tex]Z = 2.04[/tex]
[tex]Z = 2.04[/tex] has a pvalue of 0.9793
X = 0.04
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.04 - 0.07}{0.0147}[/tex]
[tex]Z = -2.04[/tex]
[tex]Z = -2.04[/tex] has a pvalue of 0.0207
0.9793 - 0.0207 = 0.9586
95.86% probability that the sample proportion will differ from the population proportion by less than 3%
