Question

high school students across the nation compete in a financial capability challenge each year by taking a financial capability exam. students who score in the top 24% are recognized publicly. assuming a normal distribution, how many standard deviations about the mean does a student have to score to be publicly recognized? (rounding to 2 decimal places)

Answer 1

Top 24% means that you have to be better than 100 - 24 = 76%.

In decimal, it is

76/100 = 0.76

This is the z-score of a normal distribution.

The formula for z-score is:

$z=\frac{x-\mu}{\sigma}$

Where

x is the raw data value

mu is the mean

sigma is the standard deviation

Since, this is a normal distribution, the mean is 0 and standard deviation 1. Also, z-score is 0.76. From z-score table, we have:

[tex]\begin{gathered} P(xRoundiing to 2 decimal places, that is 0.71 standard deviations