Question

On a test of marital intimacy husband's scores are normally distributed with a mean of 140 and a standard deviation of 25. A researcher measures 1000 husbands in the Bay Area. Using the normal curve approximation rules, approximately how many pf these husbands will have all of the above 190, below 165, and below 115? Illustrate your answer with the sketch.

Answer 1

Part a)

X ~ N ( µ = 140 , σ = 25 )
P ( X > 190 ) = 1 - P ( X < 190 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 190 - 140 ) / 25
Z = 2
P ( ( X - µ ) / σ ) > ( 190 - 140 ) / 25 )
P ( Z > 2 )
P ( X > 190 ) = 1 - P ( Z < 2 )
P ( X > 190 ) = 1 - 0.9772
P ( X > 190 ) = 0.0228

1000 * 0.0228 = 22.7501 23

Part b)

X ~ N ( µ = 140 , σ = 25 )
P ( X < 165 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 165 - 140 ) / 25
Z = 1
P ( ( X - µ ) / σ ) < ( 165 - 140 ) / 25 )
P ( X < 165 ) = P ( Z < 1 )
P ( X < 165 ) = 0.8413

1000 *0.8413 = 841.3447    841

Part c)

X ~ N ( µ = 140 , σ = 25 )
P ( X < 115 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 115 - 140 ) / 25
Z = -1
P ( ( X - µ ) / σ ) < ( 115 - 140 ) / 25 )
P ( X < 115 ) = P ( Z < -1 )
P ( X < 115 ) = 0.1587

1000 * 0.1587 = 158.6553   159