Question

Determine the range of values containing 90% of the population of x. From a very large data set (N > 50,000), x is found to have a mean value of 192.0 units with a standard deviation of 10 units. Assume x is normally distributed.

Answer 1

Answer:

The range of values containing 90% of the population of x is from 175.55 units to 208.45 units.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\u = 192, \sigma = 10$

Determine the range of values containing 90% of the population of x.

The lower end of this interval is the value of X when Z has a pvalue of 0.5 - 0.9/2 = 0.05

The upper end of this interval is the value of X when Z has a pvalue of 0.5 + 0.7/2 = 0.95

Lower end

X when Z has a pvalue of 0.05, so X when $Z = -1.645$

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

$-1.645 = \frac{X - 192}{10}$

$X - 192 = -1.645*10$

$X = 175.55$

Upper end

X when Z has a pvalue of 0.95, so X when $Z = 1.645$

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

$1.645 = \frac{X - 192}{10}$

$X - 192 = 1.645*10$

$X = 208.45$

The range of values containing 90% of the population of x is from 175.55 units to 208.45 units.