Question

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.1. (Round your answers to four decimal places.) (a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 8 pins is at least 51?

Answer 1

Answer: 0.0051

Step-by-step explanation:

Given: Mean : $\mu = 50\text{ inch}$

Standard deviation : $\sigma =1.1\text{ inch}$

Sample size : $n=8$

The formula to calculate z is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x= 51

$z=\dfrac{51-50}{\dfrac{1.1}{\sqrt{8}}}=2.57129738613\approx2.57$

The P Value =$P(Z>51)=P(z>2.57)=1-P(z<2.57)=1-0.994915=0.005085\approx0.0051$

Hence, the probability that the sample mean hardness for a random sample of 8 pins is at least 51 =0.0051