Question

The mean intelligence quotient (IQ) score is 100, with a standard deviation of 15, and the scores are normally distributed. Given this information, the approximate percentage of the population with an IQ greater than 130 is closest to _________.

Answer 1

Answer:

2.28% is the approximate percentage of the population with an IQ greater than 130.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 15

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(IQ greater than 130)

$P( x > 130) = P( z > \displaystyle\frac{130 - 100}{15}) = P(z > 2)$

$= 1 - P(z \leq 2)$

Calculation the value from standard normal z table, we have,  

$P(x > 130) = 1 - 0.9772 = 0.0228 = 2.28\%$

2.28% is the approximate percentage of the population with an IQ greater than 130.

Answer 2

The approximate percentage of the population with an IQ greater than 130 will be 2.28%.

What is a normal distribution?

The Gaussian Distribution is another name for it. The most significant continuous probability distribution is this one. Because the curve resembles a bell, it is also known as a bell curve.

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

The mean intelligence quotient (IQ) score is 100, with a standard deviation of 15, and the scores are normally distributed.

Then the approximate percentage of the population with an IQ greater than 130 will be

The value of z-score

z = (130 - 100) / 15

z = 30 / 15

z = 2

Then the probability will be

P(x > 130) = P(z > 2)

P(x > 130) = 1 - P(z < 2)

P(x > 130) = 1 - 0.97725

P(x > 130) = 0.02275

Then the percentage will be

P = 0.02275 x 100

P = 2.275%

P ≅ 2.28%

More about the normal distribution link is given below.

https://brainly.com/question/12421652

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