Answer:
C) 0.0918
Step-by-step explanation:
Let [tex]\bar{X}[/tex] be the mean pH measurement of these 4 specimens, we know that this random variable is normally distributed with mean 8 and standard deviation [tex]0.3/\sqrt{4}=0.3/2=0.15[/tex], this because we have a sample size n = 4. We are looking for [tex]P(\bar{X}>8.2)[/tex], the z-score related to 8.2 is (8.2-8)/0.15=1.3333, therefore, [tex]P(\bar{X}>8.2) = P(Z > 1.3333) = 0.0912[/tex].
