Respuesta :
Answer:
18.67% probability that the proportion of students that receive an a is 0.20 or less.
Step-by-step explanation:
Problems of normally distributed samples can be 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]\mu = 0.25[/tex].
The standard deviation of a proportion [tex]\pi[/tex] is given by the following formula.
[tex]\sigma = \sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
A university administrator expects that 25% of students in a core course will receive an a. There are 60 students. So [tex]\pi = 0.25, n = 60[/tex] and [tex]\sigma = \sqrt{\frac{0.25*(0.75)}{60}} = 0.0559[/tex]
The probability that the proportion of students that receive an a is 0.20 or less is
This is the pvalue of Z when X = 0.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.2 - 0.25}{0.0559}[/tex]
[tex]Z = -0.89[/tex]
[tex]Z = -0.89[/tex] has a pvalue of 0.1867.
So there is an 18.67% probability that the proportion of students that receive an a is 0.20 or less.
