Question

scores on the wechsler adult intelligence scale for the 20 to 34 age group are approximately normally distributed with mean 110 and standard deviation 15. how high must a person score to be in the top 5% of all scores? enter your answer as a whole number.

Answer 1

The person must score 136 to be in the top 5% of all scores in wechsler adult intelligence scale.

What is defined as the normal distribution?

• A normal distribution is a data set arrangement in which the majority of values cluster inside the center of the range and the remainder taper off symmetrically toward any extreme.
• A histogram inside a normal distribution curve is sometimes used to design the curve.

The formula for the normal distribution is;

z = (x - μ)/σ

where,

• z = z- score, taken fro table
• Mean μ = 110
• Standard deviation σ = 15
• Sample mean x.

If we want to be in the top 5%, we must outperform 95% of the remaining scores. So we must investigate.

In with us standard normal probability table, look up 0.95 and get the Z score that corresponds to that.

z = 1.7

Put the value in formula ad find x.

1. 7 = (x - 110)/15

x = 25.5 + 110

x = 135.5

x = 136 (whole number)

Thus, the person must score 136 to be in the top 5% of all scores in wechsler adult intelligence scale.

To know more about the normal distribution, here

https://brainly.com/question/23418254

#SPJ4