Question

In a study conducted in 1980, researchers calculated the height difference between couples with kids in the UK. They found that the height difference between husband and wife follows a normal distribution with mean 5.6 inches and standard deviation equal to 3.7 inches.

1) What is the difference in height for the middle 68% of the couples in the study?

2) Calculate the percentage of couples in which the man is 5 inches taller than the woman. (you have to use Table B in the book)

3) (harder) Calculate the percentage of couples that have about the same height. Assume that a couple has the same height when the height difference is less than one inch.

Answer 1

1)since 68% of the values are within 1 standard deviaiton from the mean

corresponding interval = 5.6 -/+ 3.7 or between 1.90 to 9.30

2)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	5.6
std deviation   =σ=	3.700

probability =P(X>5)=P(Z>(5-5.6)/3.7)=P(Z>-0.16)=1-P(Z<-0.16)=1-0.4364=0.5636~ 56.36%

3)

probability =P(-1<X<1)=P((-1-5.6)/3.7)<Z<(1-5.6)/3.7)=P(-1.78<Z<-1.24)=0.1075-0.0375=0.07~ 7.00%