Question

The stiffness of a certain type of steel beam used in building construction has mean 30 kN/ mm and standard deviation 2kN / m m.a. Is it possible to compute the probability that the stiffness of a randomly chosen beam is greater than 32 kN/mm? If so,compute the probability. If not, explain why not.b. In a sample of 100 beams, is it possible to compute the probability that the sample mean stiffness of the beams isgreater than 30.2 kN/mm? If so, compute the probability. If not, explain why not.

Answer 1

a) We don't know the distribution of the stiffnness of the type of steel beam, so it is not possible to compute the stiffness of a randomly chosen beam is greater than 32 kN/mm.

b) 15.87% probability that the sample mean stiffness of the beams isgreater than 30.2.

Explanation:

To solve this problem, we have 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 = (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, of at least 30, can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s= (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 30, \sigma = 2`

a. Is it possible to compute the probability that the stiffness of a randomly chosen beam is greater than 32 kN/mm? If so,compute the probability. If not, explain why not

We don't know the distribution of the stiffnness of the type of steel beam, so it is not possible to compute the stiffness of a randomly chosen beam is greater than 32 kN/mm.

b. In a sample of 100 beams, is it possible to compute the probability that the sample mean stiffness of the beams isgreater than 30.2 kN/mm? If so, compute the probability. If not, explain why not.

We do not know the distribution of the stiffnness of the type of steel beam. However, for a sample of 100 beams, by the Central Limit Theorem, we know that the distribution is going to be approximately normal.

Here we have
`n = 100, s = (2)/(√(100)) = 0.2`

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

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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 sample mean stiffness of the beams isgreater than 30.2.