Respuesta :
Answer:
[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples
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.
Z scores below -2 are considered unusually low.
Z scores above 2 are considered unusually high.
For a sample 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]\mu = 0.42, \sigma = \sqrt{\frac{0.42*0.58}{200}} = 0.0349[/tex]
Is it unusual to see 32% who rarely or never use academic advising services in one of these samples
What is the z-score for X = 0.32?
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.32 - 0.42}{0.0349}[/tex]
[tex]Z = -2.865[/tex]
[tex]Z = -2.865[/tex] means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples