Question

If I sample 50 people at random, how many people would I expect to have IQ scores… (round to the nearest person) Dtandard deviation 16 mean of standard 100

(a) below 90?

(b) below 110?

(c ) above 70

(d) above 115

Answer 1

a)

X ~ N ( µ = 100 , σ = 16 )
P ( X < 90 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 90 - 100 ) / 16
Z = -0.625
P ( ( X - µ ) / σ ) < ( 90 - 100 ) / 16 )
P ( X < 90 ) = P ( Z < -0.625 )
P ( X < 90 ) = 0.266

Of the 50 people, we expect 50 * 0.266 = 13.3 = 13 people

b)

X ~ N ( µ = 100 , σ = 16 )
P ( X < 110 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 110 - 100 ) / 16
Z = 0.625
P ( ( X - µ ) / σ ) < ( 110 - 100 ) / 16 )
P ( X < 110 ) = P ( Z < 0.625 )
P ( X < 110 ) = 0.734

Of the 50 people, we expect 50 * 0.734 = 36.7 = 37 people

c)

X ~ N ( µ = 100 , σ = 16 )
P ( X > 70 ) = 1 - P ( X < 70 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 70 - 100 ) / 16
Z = -1.875
P ( ( X - µ ) / σ ) > ( 70 - 100 ) / 16 )
P ( Z > -1.875 )
P ( X > 70 ) = 1 - P ( Z < -1.875 )
P ( X > 70 ) = 1 - 0.0304
P ( X > 70 ) = 0.9696

Of the 50 people, we expect 50 * 0.9696 = 48.48 = 48 people

d)

X ~ N ( µ = 100 , σ = 16 )
P ( X > 115 ) = 1 - P ( X < 115 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 115 - 100 ) / 16
Z = 0.9375
P ( ( X - µ ) / σ ) > ( 115 - 100 ) / 16 )
P ( Z > 0.9375 )
P ( X > 115 ) = 1 - P ( Z < 0.9375 )
P ( X > 115 ) = 1 - 0.8257
P ( X > 115 ) = 0.1743

Of the 50 people, we expect 50 * 0.1743 = 8.715 = 9people