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?

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

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

By the Central Limit Theorem

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

$30.32

step-by-step explanation:

65.00*.12=2.80

65.00-2.80=57.20

57.20*1.06=60.632

60.632/2=30.316

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1