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 13 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:

A student have to score 1.28 standard deviations above the mean o be publicly recognized.

Step-by-step explanation:

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-variates is known as the standard normal distribution.

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean.

Thus, z-scores are a type of standardized scores.

It is provided that the students who score in the top 13 percent in the National Financial Capability Challenge Exam are recognized publicly for their achievement by the Department of the Treasury.

Let x be score of the top 13%.

So, to be are recognized publicly for their achievement by the Department of the Treasury the probability of score x must be:

P (X < x) = 0.90

⇒ P (Z < z) = 0.90

The value of z for the above probability is:

z = 1.28

*Use a z-table.

Thus, a student have to score 1.28 standard deviations above the mean o be publicly recognized.

Answer 2

Answer:

The students have to score 1.13 standard deviations above the mean score to be publicly recognized.

Step-by-step explanation:

Mean = not given

Standard deviation = not given

Let Probability of Students who score in the top 13 percent = 13%

Probability of Students who score below the top 13 percent = Pr(<13%)

= 100% - 13% = 87%

Pr(<13%) = 0.87

To solve this, Use inverse of the standard normal cumulative distribution which has a mean of 0 and standard deviation of 1

Mean = 0

Standard deviation = 1

Probability used = 0.87

The z score associated with the highest 13 percent is found using excel formula:

=NORM.S.INV(0.87) = 1.1264

The students who score in the top 13 percent score is approximately 1.13 standard deviations above the mean score.

= mean + 1.13 Standard deviation