Answer: 0.0038
Step-by-step explanation:
Given : Mean : [tex]\mu=\text{60 minutes}[/tex]
Standard deviation : [tex]\sigma = \text{9 minutes}[/tex]
Sample size = 36
The formula for z -score :
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 56 ,
[tex]z=\dfrac{56-60}{\dfrac{9}{\sqrt{36}}}=-2.67[/tex]
The p-value = [tex]P(z<-2.67)[/tex]
[tex]=0.0037925\approx0.0038[/tex]
Hence, the probability that the sample mean of the sampled students is less than 56 minutes =0.0038.
