Respuesta :
Answer:
38.46% probability that the sample proportion will NOT be between 0.60 and 0.73
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 = 0.7, n = 50[/tex]
So
[tex]\mu = 0.7, \sigma = \sqrt{\frac{0.7*0.3}{50}} = 0.0648[/tex]
What is the probability that the sample proportion will NOT be between 0.60 and 0.73?
This is 1 subtracted by the probability that it is between 0.6 and 0.73.
Probability it is between 0.6 and 0.73
pvalue of Z when X = 0.73 subtracted by the pvalue of Z when X = 0.6. So
X = 0.73
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.73 - 0.7}{0.0648}[/tex]
[tex]Z = 0.46[/tex]
[tex]Z = 0.46[/tex] has a pvalue of 0.6772
X = 0.6
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.6 - 0.7}{0.0648}[/tex]
[tex]Z = -1.54[/tex]
[tex]Z = -1.54[/tex] has a pvalue of 0.0618
0.6772 - 0.0618 = 0.6154
NOT be between 0.60 and 0.73?
1 - 0.6154 = 0.3846
38.46% probability that the sample proportion will NOT be between 0.60 and 0.73
