Respuesta :
Answer:
0.9858 = 98.58% probability that the proportion of tickets sold in a sample of 741 tickets would 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 a 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, we have that:
[tex]p = 0.27, n = 741[/tex]
So
[tex]\mu = 0.27[/tex]
[tex]\sigma = \sqrt{\frac{0.27*0.73}{741}} = 0.0163[/tex]
What is the probability that the proportion of tickets sold in a sample of 741 tickets would differ from the population proportion by less than 4%.
This is the pvalue of Z when X = 0.27 + 0.04 = 0.31 subtracted by the pvalue of Z when X = 0.27 - 0.04 = 0.23. So
X = 0.31
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.31 - 0.27}{0.0163}[/tex]
[tex]Z = 2.45[/tex]
[tex]Z = 2.45[/tex] has a pvalue of 0.9929
X = 0.23
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.23 - 0.27}{0.0163}[/tex]
[tex]Z = -2.45[/tex]
[tex]Z = -2.45[/tex] has a pvalue of 0.0071
0.9929 - 0.0071 = 0.9858
0.9858 = 98.58% probability that the proportion of tickets sold in a sample of 741 tickets would differ from the population proportion by less than 4%.
