Question

Given IQ scores are approximately normally distributed with a mean of 100 and a standard deviation of 15 , the proportion of people with IQ s above 130 is: a)95% b) 68% c)5% d)2.5%​

Answer 1

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. About 68% of individuals have IQ scores in the interval 100 ± 1 ( 15 ) = [ 85 , 115 ] .

Answer 2

Answer: The proportion of people with IQ s above 130 is 2.5 %

Step-by-step explanation:

Probability density function of normal distribution curve is given by

$f(x) = \frac{1}{\beta \sqrt{2\pi } } e^{\frac{-1}{2}(\frac{x-\alpha }{\beta } )^{2} }$

f(x) ⇒ Probability density function (pdf)

α ⇒ Mean

β ⇒ Standard deviation

ATQ,

Mean = α = 100

Standard deviation = β = 15

pdf = $f(x) = \frac{1}{15 \sqrt{2\pi } } e^{\frac{-1}{2}(\frac{x-100 }{15 } )^{2} }$

Portion of people having IQ more than 130 =  $\int\frac{1}{15 \sqrt{2\pi } } e^{\frac{-1}{2}(\frac{x-100 }{15 } )^{2} }} \, dx$

Lower limit of integration = 130

Upper limit of integration ⇒ Infinity

Portion of people having IQ more than 130 = 0.025

Percentage of people having IQ more than 130 = 0.025 × 100 = 2.5 %

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