Answer: 0.008
Step-by-step explanation:
Given: Mean : [tex]\mu=1200[/tex]
Standard deviation : [tex]\sigma = 60[/tex]
Sample size : [tex]n=36[/tex]
The formula to calculate z-score is given by :_
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 1224, we have
[tex]z=\dfrac{1224-1200}{\dfrac{60}{\sqrt{36}}}=2.4[/tex]
The P-value = [tex]P(z>2.4)=1-P(z<2.4)=1-0.9918024=0.0081976\approx0.008[/tex]
Hence, the probability that the sample mean will be larger than 1224 =0.008
