Answer:
55.563
Step-by-step explanation:
Given the following :
Mean(m) point = 73
Standard deviation( sd) = 10.6
Lower 5% will not get a passing grade (those below the 5% percentile)
For a normal distribution:
The z-score is given by:
z = (X - mean) / standard deviation
5% of the class = 5/100 = 0.05
From the z - table : 0.05 falls into - 1.645 which is equal to the z - score
Substituting this value into the z-score formula to obtain the score(x) which seperates the lower 5%(0.05) from the rest of the class
z = (x - m) / sd
-1.645 = (x - 73) / 10.6
-1 645 * 10.6 = x - 73
-17.437 = x - 73
-17.437 + 73 = x
55.563 = x
Therefore, the score which seperetes the lower 5% from the rest of the class is 55.563
