Question

1. A continuous random variable may assume

   	a.	a finite number of values.
   	b.	only one value
   	c.	all values in an interval or collection of intervals
   	d.	all positive integers
   	e.	Non-numerical values such as categories or types

1 points   

QUESTION 9

1. Let F be the cumulative distribution function for a continuous random variable X. It is known at the following 5 points.

   F(0)=0

   F(1)=0.4

   F(2)=0.7

   F(3)=0.9

   F(4)=1

   Which of the following is wrong?

   	a.	Pr(1<X<3)=50%
   	b.	Pr(X<0)=100%
   	c.	Pr(X>1)=60%
   	d.	Pr(X>4)=0%
   	e.	Pr(2<X<4)=30%.

1 points   

QUESTION 10

1. During the spring, the demand for electric fans at a large home-improvement store is quite strong. The store tracks inventory to order a new shipment of fans when the inventory level falls to the "reorder point". If the demand exceeds the inventory, the store loses sales. The manager would like to reduce the probability of the shortage to 10%. (This policy is called the "90% service level" in operations management.)

   Assume that the demand is normally distributed with a mean of 1000 and a standard deviation of 200. Find the reorder point.

   	a.	1466
   	b.	1328
   	c.	1392
   	d.	1256
   	e.	1516

Answer 1

• A finite number of values is the feature of discrete random variable. So option a is not correct.
• Only a single value means that variable is not random.So option b is not correct.
• A continuous random variable may assume all values in an interval or collection of intervals. So option c is correct.
• Continuous random variable include rela numbers as well. So option d is not correct.
• A continuous random variable may assume only numerical values. So option e is not correct.

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Question 9:

Let F be the cumulative distribution function for a continuous random variable X:

F(0)=0

Probability of 0= P(0)= 0

F(1)=0.4

Probability of 1= P(1)= F(1)-F(0)= 0.4-0= 0.4

F(2)=0.7

Probability of 2= P(2)= F(2)-F(1)= 0.7-0.4= 0.3

F(3)=0.9

Probability of 3= P(3)= F(3)-F(2)= 0.9-0.7= 0.2

F(4)=1

Probability of 4= P(4)= F(4)-F(3)= 1-0.9= 0.1

Pr(1<X<3)= P(2)= 0.3= 30% (option a is incorrect)

Pr(X<0)= 0 (option b is incorrect)

Pr(X>1)= 0 (option c is incorrect)

Pr(X>4)= 1-Pr(X<4)= 1-F(4)= 1-1= 0% (Option d is correct)

Pr(2<X<4)= Pr(3)= 0.2= 20% (Option e is incorrect)

Option d is correct