Suppose that an anxiety measure has a mean of 114 and variance of 400.00. The researcher only wants to work with 56% of the lower participants. What is the largest average score the researcher should consider with a sample size of 35?

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

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