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Vilka [71]
3 years ago
10

High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capa

bility Challenge Exam. Students who score in the top 23 percent are recognized publicly for their achievement by the Department of the Treasury. Assuming a normal distribution, how many standard deviations above the mean does a student have to score to be publicly recognized

Mathematics
1 answer:
zalisa [80]3 years ago
8 0

Answer:

The students have to score 0.74 standard deviations above the mean to be publicly recognized.

Step-by-step explanation:

A random variable <em>X</em> is said to have a normal distribution with parameters <em>µ</em>    (mean) and <em>σ</em>² (variance).

If X \sim N (\mu, \sigma^{2}), then z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is, Z \sim N (0, 1).

The distribution of these <em>z</em>-scores is known as the standard normal distribution.

The <em>z</em>-score is a standardized form of the raw score, <em>X</em>. It is a numerical measurement of the relationship between a value (<em>X</em>) and the mean (<em>µ</em>) in terms of the standard deviation (<em>σ</em>). A <em>z</em>-score of -1 implies that the data value is 1 standard deviation below the mean. And a <em>z</em>-score of 1 implies that the data value is 1 standard deviation above the mean.

Let <em>X</em>  be defined as the scores of students at the National Financial Capability Challenge Exam.

It is provided that the students who score in the top 23% are recognized publicly for their achievement by the Department of the Treasury.

That is, P (X > x) = 0.23.

⇒ P(\frac{X-\mu}{\sigma}>\frac{x-\mu}{\sigma})=0.23

⇒

   P(Z>z)=0.23\\1-P(Z

The value of <em>z</em> for this probability value is:

<em>z</em> = 0.74.

*Use a <em>z</em>-table.

Thus, the students have to score 0.74 standard deviations above the mean to be publicly recognized.

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