How to get the lowest passing grade in normal distribution

Question

$$ Pr[X\le a]=0.05 $$

$$ P[{Y\le {{a-64}\over 7.1}]} = 0.05 $$

I tried up to here but I don't know what to do now... Hint please.

Answer 1

You need to find the $5$ percentile by first standardizing your variable. A standard normal table should help,http://www.mathsisfun.com/data/standard-normal-distribution-table.html , and see that the 5th percentile and 90th percentile are given by $z=1.28$ and $z=-2.6$ respectfully.

Notice that this is the inverse of the case where we're given a $z$ value and we find the percentile. Using this table, you have a percentile, and then you look up the associated $z-$ value.

Answer 2

Don't forget that normal distributions are symmetrical! P[Y<= 64+a] = P[Y>= 64-a]. Also, you know that Pr[X<=1.65] = .95 if mean = 0 and std_dev = 1, and that Pr[X <= mean + constant*std_dev] is constant. The rest should be easy to figure out.