Question

. It is known that scores on a certain IQ test follow a normal distribution with mean 100 and standard deviation 15.  For the whole population of test-takers, what proportion of scores will be greater than 124.0?  Also, the top 3% of test-takers will have scores greater than what value?  Finally, consider a random group of 16 people who take the IQ test.  For these 16 people, what is the probability that their average (mean) IQ score will be less than 97.0?

Answer 1

Solution :

Given that,

mean = = 100

standard deviation = = 15

a ) P (x > 124.0 )

= 1 - P (x < 292 )

= 1 - P ( x -  / ) < ( 124.0 - 100 / 15)

= 1 - P ( z < 24 / 15 )

= 1 - P ( z < 1.60 )

Using z table

= 1 - 0.9452

= 0.0548

Probability = 0.0548

b ) P( Z > z) = 3%

P(Z > z) = 0.03

1 - P( Z < z) = 0.03

P(Z < z) = 1 - 0.03

P(Z < z) = 0.97

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 15+ 100

x = 128.2

c )n = 10

= 1200

= / n = 15 16= 3.75

P( x < 97.0 )

P ( x - / ) < ( 97.0 - 100 / 3.75)

P ( z < -3 /3.75 )

P ( z <0.963)

= 0.2119

Probability = 0.2119