Note the following holds true for two normal random variables: Let X and Y are i

Question

Note the following holds true for two normal random variables:

Let X and Y are independent with X N ( x , 2 x ) and Y N ( y , y ).

Then, X + Y N ( x + y , 2 x + 2 y ); X Y N ( x y , 2 x + 2 y )

Now, define two independent random variables as follows:

{ X : the amount of chocolate that a male student in STAT509 had on Valentine’s Day N (120, 20 2 )

Y : the amount of chocolate that a female in STAT509 student had on Valentine’s Day N (80, 10 2 ) where the unit is in gram.

Suppose a male and a female students are chosen independently and at random from our STAT509 class. Find out the following probabilities using z-table.

Please provide the full procedures.

1. What is the probability that the male student had at most 90 grams of chocolate?

2. What is the probability that the female student had at least 90 grams of chocolate?

3. What is the probability that the two student had less than 180 of chocolate in total?

4. What is the probability that the male student would have at least 15 grams more than the female student?

Explanation / Answer

#1.
mean = 120
std. dev. = 20

P(X <= 90)
= P(z <= (90 - 120)/20)
= P(z <= -1.5)
= 0.0668 (using normal distribution table)

#2.
mean = 80
std. dev. = 10

P(Y >= 90)
= P(z >= (90 - 80)/20)
= P(z >= 0.5)
= 0.3085 (using normal distribution table)

#3.
mean of X+Y = 120 + 80 = 200
std. dev. of X+Y = sqrt(20^2 + 10^2) = 22.3607

P(X+Y < 180)
= P(z < (180 - 200)/22.3607)
= P(z < -0.8944)
= 0.1855

#4.
mean of X-Y = 120 - 80 = 40
std. dev. of X-Y = sqrt(20^2 + 10^2) = 22.3607

P(X-Y >= 15)
= P(z >= (15 - 40)/22.3607)
= P(z >= -1.118)
= 0.8682

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