Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.3.

(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 14 pins is at least 51?

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 35 pins is at least 51?

Answer :

Given :

The population mean $(\mu )=50$

The population standard deviation $(\sigma )=1.3$

Solution :

(A) The Probability of Sample Mean Hardness At Least 51 for 14 Pins :

$$\therefore \text{P}(\bar{x}>51)=\text{P}(\frac{\bar{x}–\mu }{\sigma /\sqrt{n}}>\frac{51–50}{\frac{1.3}{\sqrt{14}}})$$ $$=\text{P}(z>2.878)$$ $$=0.0020$$

(B) The Probability of Sample Mean Hardness At Least 51 for 35 Pins :

$$\therefore \text{P}(\bar{x}>51)=\text{P}(\frac{\bar{x}–\mu }{\sigma /\sqrt{n}}>\frac{51–50}{\frac{1.3}{\sqrt{35}}})$$ $$=\text{P}(z>4.551)$$ $$=0.0000$$