Question

The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 503 MPa with a standard deviation of 13 MPa. (a) What is the probability that a randomly chosen sample of glass will break at less than 503 MPa?

Answer 1

Answer:

0.50 or 50%

Explanation:

Given:

Mean of fracture strength (μ) = 503 MPa

Standard deviation (σ) = 13 MPa

Strength of the given sample (x) = 503 MPa

The distribution is normal distribution.

So, first, we will find the z-score of the distribution using the formula:

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

Plug in the values and solve for 'z'. This gives,

$z=\frac{503-503}{13}=0$

So, the z-score of the distribution is 0.

Now, we need the probability P(z < 0).

We know that, z = 0 is the middle point of a normal distribution curve. So, the point z = 0 divides the area under the curve into 2 equal halves.

Therefore, P(z < 0) = 0.50 = 50%

Hence, the probability that a randomly chosen sample of glass will break at less than 503 MPa is 0.50 or 50%.