Question

On a planet far far away from Earth, IQ of the ruling species is normally distributed with a mean of 119 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. a. What is the distribution of X? X ~ N( 119 , 15 ) b. Find the probability that a randomly selected person's IQ is over 113. Round your answer to 4 decimal places. c. A school offers special services for all children in the bottom 2% for IQ scores. What is the highest IQ score a child can have and still receive special services? Round your answer to 2 decimal places. d. Find the Inter Quartile Range (IQR) for IQ scores. Round your answers to 2 decimal places. Q1: Q3: IQR:

Answer 1

Solution :

Given that ,

mean = = 119

standard deviation = = 15

a) The distribution of x is normal X ~ N( 119 , 15)

b) P(x > 113 ) = 1 - p( x< 113 )

=1- p P[(x - ) / < (113 - 119) / 15]

=1- P(z < -0.40 )

Using z table,

= 1 - 0.3446

= 0.6554

c) Using standard normal table,

P(Z < z) = 2%

= P(Z < z) = 0.02  

= P(Z < -2.054) = 0.02

z = -2.054

Using z-score formula,

x = z * +

x = -2.054 * 15 * 119

x = 88.19

d) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 15 + 119

x = 108.88

First quartile =Q1 = 108.88

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.6745 ) = 0.75

z = 0.6745

Using z-score formula,

x = z * +

x = 0.6745 * 15 + 119

x = 129.12

Third quartile =Q3 = 129.12

IQR = Q3 - Q1

IQR = 129.12 - 108.88

IQR = 20.24