Mathematics, 06.05.2020 02:22 mruffier6239

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

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Answer from: spobyyy

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.

$P(Xx)=0.23\\\\P(Zz)=0.23\\\\1-P(Z$

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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High school students across the nation compete in a financial capability challenge each year by taki

Answer from: mkim2290

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.

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

⇒

$P(Zz)=0.23\\1-P(Z$

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 from: Quest

c.35

step-by-step explanation:

8+13=21

21+8=29

29+6=35

Answer from: Quest

1. 5/82.95% is better than 7/8

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