# Math Help - Probability.

1. ## Probability.

In some cities tall people who want to meet and socialize with other tall people can join Beanstalk Clubs. To qualify, a man must be over 6'2" and a woman over 5'10." According to the National Health Survey, heights of adults may have a Normal model with mean heights of 69.1" for men and 64" for women. The respective standard deviations are 2.8" and 2.5."

a) You're probably not surprised to learn that men are generally taller than women, but what does the greater standard deviation for men's heights indicate?
A greater standard deviation for men indicates that the amount of men with heights closest to the mean is less than the amount of women closest to the mean. It indicates a greater variation in height for men.

b) Are men or women more likely to qualify for Beanstalk membership?
men

c) Beanstalk members believe that height is an important factor when people select their spouses. To investigate, we select at random a married man and, independently, a married woman. Define two random variables, and use them to express how many inches taller the man is than the woman.
I don't really understand the directions....am I supposed to just make up two random variables, and how would they influence the heights??

d) What's the mean of this defference?

e) What's the standard deviation of this difference?

f) What's the probability that the man is taller than the woman (that the difference in heights is greater than 0)?

g) Suppose a survey of married people reveals that 92% of husbands were taller than their wives. Based on your answer to part f, do you believe that people chose spouses independent of height?
I don't know the answer to f, but I would say that people do not choose their spouses independent of height.

2. (a) looks ok

(b) You need to compare these two calculations...

$P(M>74 inches)=P\biggl(Z>{74-69.1\over 2.8}\biggr)$

and

$P(W>70 inches)=P\biggl(Z>{70-64\over 2.5}\biggr)$

3. Man-Woman (in height) is a normal random variable
with mean 69.1-64=5.1 inches and variance $(2.8)^2+(2.5)^2$

$P(man-woman>0)=P\biggl(Z>{0-5.1\over\sqrt{(2.8)^2+(2.5)^2 }}\biggr)$

4. Originally Posted by matheagle
Man-Woman (in height) is a normal random variable
with mean 69.1-64=5.1 inches and variance $(2.8)^2+(2.5)^2$

$P(man-woman>0)=P\biggl(Z>{0-5.1\over\sqrt{(2.8)^2+(2.5)^2 }}\biggr)$
I'm a little confused...if z stands for z-score, which I am assuming it does, then the right calculation (which calculates the z-score) would be z, so how could z be greater than it.....?

5. Originally Posted by samipon
I'm a little confused...if z stands for z-score, which I am assuming it does, then the right calculation (which calculates the z-score) would be z, so how could z be greater than it.....?
It means that you're meant to calculate the probability that Z is larger than that number.

6. figure out that number and look up the probability on a Z table
for example P(Z>0)=.5