Question

The stiffness of a certain type of steel beam used in building construction has mean 30 kN/mm and standard deviation 2 kN/mm. What is the probability that the average of 100 randomly selected beams is greater than 30.2 kN/mm?

Answer 1

Answer:

15.87% probability that the average of 100 randomly selected beams is greater than 30.2 kN/mm

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 30, \sigma = 2, n = 100, s = \frac{2}{\sqrt{100}} = 0.2$

What is the probability that the average of 100 randomly selected beams is greater than 30.2 kN/mm?

This probability is 1 subtracted by the pvalue of Z when X = 30.2. So

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

By the Central Limit Theorem

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

$Z = \frac{30.2 - 30}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

1 - 0.8413 = 0.1587

15.87% probability that the average of 100 randomly selected beams is greater than 30.2 kN/mm