Respuesta :
Answer:
0.0228 = 2.28% probability that the proportion of tickets sold in a sample of 592 tickets would be less than 13%
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]p = 0.16, n = 592[/tex]
So
[tex]\mu = 0.16, \sigma = \sqrt{\frac{0.16*0.84}{592}} = 0.015[/tex]
What is the probability that the proportion of tickets sold in a sample of 592 tickets would be less than 13%?
This is the pvalue of Z when X = 0.13. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.13 - 0.16}{0.015}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
0.0228 = 2.28% probability that the proportion of tickets sold in a sample of 592 tickets would be less than 13%
