Question

5.In a certain Asian country, the average weight of men is 60 kg and its standard deviation is 8 kg. The average of women is 52 kg and its standard deviation is 5 kg. If we select one man and one woman randomly, (5.1) What is the probability that the sum exceeds 100 kg? (5.2) What is the probability that the woman's weight exceeds the man's weight? (53) What is the probability that the woman's weight is at least ¾ of the man's weight?

Please use the following distribution table to get the answer!

Answer 1

5:

Let X shows the man weight and Y shows the woman weight. Here we have

$mu_{X}=60,sigma_{X}=8,mu_{Y}=52,sigma_{Y}=5$

1:

Let T= X+Y

Here T will have normal distribution with mean

$mu_{T}=mu_{X}+mu_{Y}=60+52=112$

and standard deviation

$sigma_{T}=sqrt{sigma^{2}_{X}+sigma^{2}_{Y}}=sqrt{64+25}=9.434$

The z-score for T = 100 is

T HT 100 - 112 9.434

The required probability is

P(T > 100) = P(z > -1.27) = 1 - P(z <= -1.27) = 1- 0.102 = 0.8980

(2)

Let D =X-Y

Here D will have normal distribution with mean

$mu_{D}=mu_{X}-mu_{Y}=60-52=8$

and standard deviation

$sigma_{D}=sqrt{sigma^{2}_{X}+sigma^{2}_{Y}}=sqrt{64+25}=9.434$

The z-score for D = 0 is

The required probability is

P(D < 0) = P(z < -0.85) = 0.1977

(3)

Let W =(3X /4)-Y

Here W will have normal distribution with mean

$mu_{W}=rac{3mu_{X}}{4}-mu_{Y}=-7$

and standard deviation

The z-score for W = 0 is

The required probability is