Question

The strength of an aluminum alloy is normally distributed with mean 10 gigapascals (GPa) and standard deviation 1.4 GPa. What is the first [lower] quartile of the strengths of this alloy

Answer 1

Answer:

The first quartile of the strengths of this alloy is 9.055 GPa.

Step-by-step explanation:

Problems of normal distributions 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.

The strength of an aluminum alloy is normally distributed with mean 10 gigapascals (GPa) and standard deviation 1.4 GPa.

This means that $\mu = 10, \sigma = 1.4$

What is the first [lower] quartile of the strengths of this alloy?

This is the 100/4 = 25th percentile, which is X when Z has a pvalue of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 10}{1.4}$

$X - 10 = -0.675*1.4$

$X = 9.055$

The first quartile of the strengths of this alloy is 9.055 GPa.