Question

Dear Math Helper,

Answer 1

Answer:

n = 98, that is, she scored at the 98th percentile.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

She scored 38, so $X = 38$

Test scores are normally distributed with a mean of 25 and a standard deviation of 6.4.

This means that $\mu = 25, \sigma = 6.4$

Find the percentile:

We have to find the pvalue of Z. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{38 - 25}{6.4}$

$Z = 2.03$

$Z = 2.03$ has a pvalue of 0.98(rounding to two decimal places).

So n = 98, that is, she scored at the 98th percentile.