Probability of score greater than 40 by Normal Distribution analysis Part A is straight forward. Part B am not able to do stuck with the thought that a score (value)X has to be give to John for z= X-mean/std. Pointers would be helpful
There are 20 true-false questions. Each question is worth 5 points Assume that John has no idea about the questions and randomly guess all questions
a. What is the probability that he gets at least the passing mark of 60 points? 
b. Assume that the score of the students follow a normal distribution with the mean being 60 points and standard deviation being 5 points, what is the probability that John scores higher than 40% of students? (Hint: Think about the normal distribution bell curve, 40% is how many sigma away from the mean?)
 A: He needs to do better than $40\%$ of students. That's $10\%$ of students away from the mean. Using this Z-table we see that the closest $Z$ for $0.1$ is $0.25$. If a more precise number is desired, a calculator such as this one can be used to get the value of $0.253347$, though for this particular case $0.25$ works fine. Since it's to the left of the mean, we subtract the score difference from the mean.
$$60-0.25\cdot5=58.75$$
So to do better than $40\%$ of students, John needs to get $58.75$ points. Since points come in increments of $5$, that can't happen. You either need to go down to $55$, in which case John does better than $\approx 15.8655\%$ of students, or you go up to $60$ and John does better than $50\%$ of students. That one is closer to $40\%$, so that's what I'd go with. It also follows with John needing to do better than at least $40\%$ of students, which unlike exactly $40\%$ is calculable. As such John needs $60$ points. This makes part B like part A. I won't explain that part since you said it was straight forward, though I can if needed.
