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jenyasd209 [6]
3 years ago
6

A university administrator expects that 25% of students in a core course will receive an

Mathematics
2 answers:
Naya [18.7K]3 years ago
8 0

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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:

\mu = 0.25.

The standard deviation of a proportion \pi is given by the following formula.

\sigma = \sqrt{\frac{\pi(1-\pi)}{n}}

A university administrator expects that 25% of students in a core course will receive an a. There are 60 students. So \pi = 0.25, n = 60 and \sigma = \sqrt{\frac{0.25*(0.75)}{60}} = 0.0559

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

Z = \frac{X - \mu}{\sigma}

Z = \frac{0.2 - 0.25}{0.0559}

Z = -0.89

Z = -0.89 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.

xxMikexx [17]3 years ago
3 0
75% i would think if you subtract 25% from 100% of the class
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