Answer: the cutoff score to get into the class is 43
Explanation:
Since the scores on the test are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = scores on the test.
µ = mean score
σ = standard deviation
From the information given,
µ = 38
σ = 6
The probability value for the 80th percentile is 80/100 = 0.8
Looking at the normal distribution table, the z score corresponding to the probability value is 0.85
Therefore,
0.85 = (x - 38)/6
Cross multiplying by 6, it becomes
0.85 × 6 = x - 38
5.1 = x - 38
x = 5.1 + 38
x = 43.1
Approximately 43