Question

A distribution of values is normal with a mean of 195.2 and a standard deviation of 68.7. Find P6, which is the score separating the bottom 6% from the top 94%.

Answer 1

Answer:

P6 = 88.3715

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 195.2, \sigma = 68.7$

Find P6, which is the score separating the bottom 6% from the top 94%.

This is the value of X when Z has a pvalue of 0.06. So it is X when Z = -1.555.

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

$-1.555 = \frac{X - 195.2}{68.7}$

$X - 195.2 = -1.555*68.7$

$X = 88.3715$

P6 = 88.3715