Which of the following tables shows a valid probability density function? a. x P(X=x) 0 38 1 14 2 38 b. x P(X=x) 0 0.2 1 0.1 2 0.35 3 0.17 c. x P(X=x) 0 910 1 −310 2 310 3 110 d. x P(X=x) 0 0.06 1 0.01 2 0.07 3 0.86 e. x P(X=x) 0 12 1 18 2 14 3 18 f. x P(X=x) 0 110 1 110 2 310 3

Step-by-step explanation:

Since we know that for a distribution be a probability density function sum of all the probability events should be equal to 1 and all individual events should have probability between 0 and 1

a. x P(X=x)

0 3/8

1 1/4

2 3/8

P(X=0)+P(X=1)+P(X=2) = 3/8 + 1/4 + 3/8

P(X=0)+P(X=1)+P(X=2) = 6/8 + 2/8 = 1

This is a probability density function

b. x P(X=x)

0 0.2

1 0.1

2 0.35

3 0.17

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.2 + 0.1 + 0.35 + 0.17

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.65 + 0.17 = 0.82 ≠ 1

Therefore this is NOT a probability density function

c. x P(X=x)

0 9/10

1 −3/10

2 3/10

3 1/10

Since P(X=1) is not between 0 and 1

Therefore this is NOT a probability density function

d. x P(X=x)

0 0.06

1 0.01

2 0.07

3 0.86

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.06 + 0.01 + 0.07 + 0.86

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.14 + 0.86 = 1

Therefore this is a probability density function

e. x P(X=x)

0 1/2

1 1/8

2 1/4

3 1/8

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/2 + 1/8 + 1/4 + 1/8

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/2 + 1/2 = 1

Therefore this is a probability density function

f. x P(X=x)

0 1/10

1 1/10

2 3/10

3 1/5

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/10 + 1/10 + 3/10 + 1/5

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 2/10 + 5/10 = 7/10 ≠ 1

Therefore this is NOT a probability density function

Let’s simplify 1 1/2 cups of baking soda to 3/2. And you have 1/2 cups of conditioner.

If you need 1/2 cups of baking soda you need to figure out how many times 1/2 fits into 3/2. 3 times.

So you need a third of the conditioner.
1/3 x 1/2=1/6
1/6 cups of conditioner