Question

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

Answer 1

Answer:

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

Step-by-step explanation:

A random variable X is said to have a normal distribution with parameters µ    (mean) and σ² (variance).

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

The distribution of these z-scores is known as the standard normal distribution.

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

Let X  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.

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

⇒

   $P(Z>z)=0.23\\1-P(Z<z)=0.23\\P(Z<z)=0.77$

The value of z for this probability value is:

z = 0.74.

*Use a z-table.

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

Answer 2

A standard deviation is a measure of how dispersed the data is in relation to the mean.

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

• A normal distribution is the proper term for a probability bell curve.

• In a normal distribution the mean is zero and the standard deviation is 1.

• It has zero skew and a kurtosis of 3.

Let X represent that students who score in the top 23 % and  recognized publicly.

    We know that,   $z=\frac{x-\mu}{\sigma}$ , where $\mu$ is mean and $\sigma$ is standard deviation.

  $P(X>x)=0.23\\\\P(Z>z)=0.23\\\\1-P(Z<z)=0.23\\\\P(Z<z)=1-0.23=0.77$

In the z - table, value of z corresponding to probability 0.77 is 0.74.

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

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