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A genius society requires an IQ that is in the top 2% of the population in order to join. If an IQ test has a mean of 100 and a standard deviation of 15, which of the following could be used as the minimum qualifying score to join the genius society?

Respuesta :

i just did the test 130 is correct


The minimum qualifying score to join the genius society is 130.8

What is nominal distribution?

A probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean is called nominal distribution or Gaussian distribution

What is mean?

The mean is the arithmetic average of a set of given numbers. In nominal distribution the mean is the central tendency of the normal distribution. It defines the location of the peak for the bell curve.

What is standard deviation?

The standard deviation is a measure of variability. It defines the width of the normal distribution. The standard deviation determines how far away from the mean the values tend to fall. It represents the typical distance between the observations and the average.

From the question, mean(μ)  = 100 and standard deviation(σ)= 15

The standard normal distance from the mean, z, is given by

z=(x−μ)/σ

Where x is a particular measurement, μ is the population mean, and σ is the population standard deviation.

According to the question

probability of IQ of top 2% in the population is  Pr(X > x) = 0.02

probability of IQ less then top 2% in the population is  pr(X<x) = 0.98

we have to determine z value using the formula z=(x−μ)/σ and z table

(0.98-100)/15 =-3.93

from z table the z value is 2.05

So, the minimum qualifying score to join the genius society is

2.05=(x- 100)/15

x= 130.8

Thus the following could be used as the minimum qualifying score to join the genius society is 130.8

To know about nominal distribution click here

brainly.com/question/20376026

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