Question

Find the standard deviation of the distribution in the following situations. n

(a) MENSA is an organization whose members have IQs in the top 1% of the population. IQs are normally distributed with mean 100, and the minimum IQ score required for admission to MENSA is 130. σ=

(b) Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 181 milligrams per deciliter (mg/dl). Women with cholesterol levels above 224 mg/dl are considered to have high cholesterol and about 17% of women fall into this category. σ=

Answer 1

(a)

X : IQ score

X follows normal distribution with mean 100 standard deviation :

Given,

MENSA is an organization whose members have IQs in the top 1% of the population.

minimum IQ score required for admission to MENSA is 130

i.e 1% of population has IQ more than 130 : P(X>130) = 1/100 =0.01

P(X>130) = 1-P(X130) = 0.01 ; P(X130) =1-0.01=0.99

Z₁ : Z-score for 130 = (130 - Mean)/standard deviation = (130-100)/ = 30/

= 30/Z₁

P(ZZ₁) = P(X130) =0.99

From standard normal tables,

P(Z2.33) =0.9901 ; Z₁ = 2.33

= 30/Z₁ = 30/2.33 = 12.8755

Answer : = 12.8755

(b)

X : Cholesterol levels for women aged 20 to 34

X follows normal distribution with mean 181 and standard deviation

Given,

Women with cholesterol levels above 224 mg/dl are considered to have high cholesterol and about 17% of women fall into this category

i.e P(X>224) = 0.17

P(X>224) = 1-P(X224) = 0.17 ; P(X224) =1-0.17 = 0.83

Z₁ : Z-score for 224= (224- Mean)/standard deviation = (224-181)/ = 43/45.2632

= 43/Z₁

P(ZZ₁) = P(X224) =0.83

From standard normal tables,

P(Z0.95) =0.8289 ; Z₁ = 0.95

= 43/Z₁ = 43/0.95 = 45.2632

Answer : = 45.2632