Question

Suppose that a wall is constructed from concrete blocks where a block is a normal random variable with mean 11 kg and standard deviation 0.09 kg. If 85% of the blocks have weights less than k kg, evaluate k

Answer 1

Answer:

The value for k is about k = 11.093kg.

Step-by-step explanation:

To answer this question, that is, P(x<k) = 0.85, we can use the standardized value for a raw score, or the formula for obtaining z-scores:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

$\\ x$ is the raw score.

$\\ \mu$ is the population mean.

$\\ \sigma$ is the population standard deviation.

In this case, we are asking to find a value x = k, so that 85% of the blocks have weights less than k (kg).

We have also to use the values for the cumulative standard normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ to find the value of z that corresponds to a probability of 0.85.

First step: find the z that corresponds to the probability of 0.85.

To use the formula [1], as we previously mentioned, we first need to find the value of z that corresponds to a probability of 0.85 using the cumulative standard normal distribution table (available in any Statistics books or on the Internet). Then, for a cumulative probability of 0.85, the corresponding value for z = 1.03 (approximately).

Second step: solve the formula [1] for x.

Solving this formula for x, we have:

$\\ z = 1.03$

$\\ \mu = 11kg$

$\\ \sigma = 0.09kg$

Then (without units):

$\\ z = \frac{x - \mu}{\sigma}$

$\\ 1.03 = \frac{x - 11}{0.09}$

$\\ 1.03*0.09 = x - 11$

$\\ (1.03*0.09) + 11 = x$

So

$\\ x = (1.03*0.09) + 11$

$\\ x = 0.0927 + 11$

$\\ x = 11.0927 \approx 11.093 = k$

Thus, this value for x = 11.093 equals the value for k (x = k) so that

$\\ P(x$

Then

$\\ P(x$ or

$\\ P(x$

We can see the graph below showing this value of k = 11.093kg for which 85% of the blocks have weights less that it (x = k = 11.093kg).