Answer:
The largest average score the researcher should consider is 117
Step-by-step explanation:
Mean = xbar = 114
Standard deviation = σ = √(variance) = √(400) = 20
Let The standardized score that corresponds with lower 56% be x'.
Let the z value that corresponds to the probability of 0.56 be z'
We need to obtain the z value that corresponds with a Probability of P(z ≤ z') = 0.56
using the normal probability distribution table,
z' = 0.151
P(z ≤ z') = P(z ≤ 0.151) = 0.56
z' = (x' - xbar)/σ
0.151 = (x' - 114)/20
x' = 117
