These problems can be solved two ways: finding the Z scores and looking up values in a table, or using technology.
1) The Z score is just the number of standard deviations above or below the mean that a particular score falls. Z scores that are negative are below the mean. Z scores that are positive are above the mean. Z = (x - mu)/sigma is the formula, where x is the value we are interested in, mu is the mean and sigma is the standard deviation. Z(65) = (65-70)/4 = -5/4 = -1.25. Looking up -1.25 in the Z table, we get a value of 0.1056. This is the probability that we get a value lower than 65. So the probability that we get a value higher than 65 is 1-0.1056 = .8944 This is our answer. Alternately, you can use the normalcdf function on your TI 84(2nd ->VARS, then option 2: normalcdf). normalcdf with a lower of 65, an upper of 100 (effectively infinity), a mean of 70, and a standard deviation of 4 gives you .89435016, which rounds to .8944, the same answer as with the Z scores.
2) Using the calculator, we get that 0.62% of the students will score below a 60. (normalcdf(-1000,60,70,4) = .0062). 0.62% of 500 is 3.10. So about 3 students are expected to score below a 60. Z(60) = (60-70)/4 = -2.5. Looking up -2.50 in the Z table gives us the same .0062. (A special note. To enter a negative number, use the (-) button, next to the . and Enter buttons, not the - grey button that's between X and +.
3) If you look up .9500 in the body of the Z table (yours might need .4500), you get a Z score of about 1.645. using the Z score formula we get (1.645 = (x - 70)/4). Solving, you get x = 76.58. So to be in the top 5%, you'd have to score above a 76.58. Using the 2nd->VARS->invNorm(.95,70,4) function on your TI 84 gives 76.579.
