Question

1. Determine the value for the constant "k" that will make the function below a PDF.

f(x) = k x^4 for x in (0,2) and f(x) = 0, for all other x ; k =

A. 8/3 B. 3/8 C. 3/64 D. 22/7 E. None of these

2. In a dice tossing random experiment , a red die and a green die are thrown independently . Consider as a random variable, X, the range in showing dots. for example if the red die is 6 and the green die is 4 , then the random variable has a value of 2 in that experiment. Calculate the probability, Pr(x=1) =

A) .278 B) .167 C) .111 D) .0556

3. Assume Z is a standard normal random variable, what is the area to right of z = -1.23

A) .1093 B) .8907 C) .3594 D) .6406 E) None of these  

Answer 1

Solution:

Question 1:

Given:

We have to find value of k such that f(x) becomes a PDF.

Thus if f(x) is a PDF then it must satisfy following property:

Since none of the option shows value = 5/32

thus correct option is e) None of these.

Question 2)

A red die and a green die are thrown independently.

X = the range in showing dots.

We have to find the probability, Pr(x=1) =..............?

Sample Space S is:

X = Range is 1

So we have to find outcomes which shows Range is 1

{ (2,1) , (1,2) , (3,2) , ( 2,3) , (4,3) , ( 3,4) , (5,4) , ( 4,5) , ( 6,5) , ( 5,6) }

These outcomes has range = 1, that is: 2-1 = 1 , 1-2 = -1, here we use absolute differences.

Thus number of outcomes whose range = 1 are = 10

that is : m = 10

and n= total outcomes in sample space = 36

Thus

Thus correct option is: A) 0.278

Question 3)

Find area to right of z = -1.23.

That is find: P( Z > -1.23) = ........?

Thus

P( Z > -1.23) = 1 - P( Z < -1.23)

Look in z table for z = -1.2 and 0.03 and find area.

P( Z < -1.23) = 0.1093

Thus

P( Z > -1.23) = 1 - P( Z < -1.23)

P( Z > -1.23) = 1 - 0.1093

P( Z > -1.23) = 0.8907

Thus correct option is: B) 0.8907