Question

The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The lowest 10 percent of the citizens would need at least how many minutes to complete the form

Answer 1

Answer:

The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.

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:

$\mu = 40, \sigma = 10$

The slowest 10 percent of the citizens would need at least how many minutes to complete the form

This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.

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

$1.28 = \frac{X - 40}{10}$

$X - 40 = 1.28*10$

$X = 52.8$

The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.