Answer: 0.5%
Step-by-step explanation:
We assume that the test scores in final exam is normally distributed.
Given : Mean :[tex]\mu= 72[/tex]
Standard deviation : [tex]\sigma=9.2[/tex]
Sample size : n= 35
Let x be the random variable that represents the test scores of the students.
The formula for the z-score :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x=76
[tex]z=\dfrac{76-72}{\dfrac{9.2}{\sqrt{35}}}\approx2.57[/tex]
By using the standard normal distribution table ,
The p-value =[tex]P(x>76)=P(z>2.57)=1-P(z\leq2.57)[/tex]
[tex]=1-0.994915=0.005085=0.5085\%\approx0.5\%[/tex]
Hence, the probability that that the mean of their test scores will be greater than 76 =0.5%
