Explanation:
we need to use the standard normal distribution table for mean = 0, standard deviation = 1.
to use it for our normal distribution with mean = 81.6 and standard deviation = 7.2 we need to find the transformation from the standard normal distribution to our normal distribution.
this is done by the z-scores of the interval limits (the z-score tells us how many standard deviations the desired value is away from the mean value).
and to correlate the given %-values to the distribution interval limits, we need to look up the corresponding p-value (probability value) in the table.
and then we need to calculate backwards to get our specific interval limits.
for the D-grade we are looking for the upper limit to "below the top 77%", which is "below the bottom 23%".
we have to assume this means "below or equal to the bottom 23%", just in case.
the lower level will be "above the bottom 8%".
so, the p-value for 23% = 0.2300.
that gives us the z-score ≈ -0.74.
z-score = (interval-limit - mean-value)/standard-deviation
-0.74 = (interval-limit - 81.6)/7.2
-0.74×7.2 = interval-limit - 81.6
-5.328 = interval-limit - 81.6
interval-limit = 76.272 ≈ 76
the p-value for 8% = 0.0800.
that gives us the z-score ≈ -1.41
-1.41 = (interval-limit - 81.6)/7.2
-1.41×7.2 = interval-limit - 81.6
-10.152 = interval-limit - 81.6
interval-limit = 71.448 ≈ 71
the students with a score s
71 < s <= 76
get a D.
(greater than 71 but lower or equal to 76).