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

choose (5, 1.5) and (-1.5,-3.5)

step-by-step explanation:

5 - 1.5 =3.5

-1.5 - (-3.5) = 2

3.5 + 2 =5.5

hope this !