Question

The fracture strength of tempered glass averages 14.1 (measured in thousands of pounds per square inch) and has standard deviation 2. (a) What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.3

Answer 1

Answer:

0.15866 is the probability of the average fracture strength of selected 100 pieces of glasses that of exceed 14.3.

Step-by-step explanation:

Given that, the strength of a tempered(x) a glass has average 14.1 and has standard deviation 2.

$\mu_x=14.1$ , $\sigma_x=2$

n=number of selected pieces= 100.

The probability that the average fracture strength of 100 pieces of this glass exceed 14.3 is

$P(\overline X>14.3)=P(\frac{\overline X-\mu_x}{2/\sqrt n}>\frac{14.3-14.1}{2/\sqrt{100}})$

                    $=P(\overline X>1)$

                    $=1-P(\overline X\leq 1)$

                    = 1 - 0.84134

                    =0.15866.