The tensile strength (ksi) of a certain type of steel at room temperature is normally distributed with a mean of 100 ksi and a standard deviation of 8.2 ksi. If X is the tensile strength of a randomly selected sample of this steel find P(X>115);
well, since this is normal distribution, I found this to be P(Z/sqrt(8.2) > 5.115) by using x-mew / sigma > 115 - mew / sigma but i can't find the value since the z table only goes up to 3.49.
any help would be appreciated
Agggh! I somehow misread the question. many thanks to pointing it out!
Note that $8,2$ is the standard desviation and not the variance. So it is
$$P(X>115)=P\left(Z=\dfrac{X-110}{8,2}>\dfrac{115-110}{8,2}\right).$$