SOLUTION: A final exam in Math 1148 has a mean of 73 with standard deviation 7.8. If a student is randomly selected, what is the probability that his/her test score is at least an 80?

 

Hi,

The 9999 is just a 'place holder' so to speak (A Number SO big, relative to the parameters given...

that the 'z' representing it...it is going to be VERY far to the right on the Normal curve 

(See below how far right 3 SDs is)

Using TI Calcuator

The syntax is normalcdf(smaller, larger, µ, σ).

Note: The 9999 is used as the larger value to be at least 5 standard deviations from the mean.

P(x≥ 80) =  normalcdf(80, 9999, 73, 7.8).

Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  

Note: z = 0 (x value the mean) 50% of the area under the curve is to the left and %50 to the right