Question

The output from a statistical computer program indicates that the mean and standard deviation of a data set consisting of 200 measurements are $1,500 and $300, respectively. Suppose the data is mound shaped and symmetrically distributed. Find the limit that 97.5% of the data points will be above.

Answer 1

Answer:

The limit that 97.5% of the data points will be above is $912.

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:

$\mu = 1500, \sigma = 300$

Find the limit that 97.5% of the data points will be above.

This is the value of X when Z has a pvalue of 1-0.975 = 0.025. So it is X when Z = -1.96.

So

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

$-1.96 = \frac{X - 1500}{300}$

$X - 1500 = -1.96*300$

$X = 912$

The limit that 97.5% of the data points will be above is $912.