Question

1. The following probability distribution represents the number of people living in a Household (X), and the probability of occurrence (P(X)). Compute the Expected Value (mean), the Variance and the Standard Deviation for this random variable. Show Your Calculations for the Mean.

    X      1         2        3          4        5               

P(X)    .33      .29      .27        .07       .04     

2. Use the binomial formula to compute the probability of a student getting 8 correct answers on a 10 question Quiz, if the probability of answering any one question correctly is 0.64. SHOW YOUR WORK.

3. Submit your answers to the following binomial questions. You may use the appendix table B #5 to answer parts (a) and (b). According to a government study, 20% of all children live in a household that has an income below the poverty level. If a random sample of 15 children is selected:

a) what is the probability that 6 or more live in poverty?

b) what is the probability that 5 live in poverty?

c) what is the expected number (mean) that live in poverty?

Answer 1

1)

Expected value :

Calculation :

Expected value = E(x) = 2.2

Variance :

Calculation :

Variance = v(x) = 1.2

Standard deviation = = = 1.0954

2)

Given ,

n = 10 , p = 0.64

We have to find P( x = 8 )

Using Binomial formula :

        { ⁿC_x = n! / x!(n-x)! }

Probability of a student getting 8 correct answers on a 10 question Quiz is 0.1642

3)

Given ,

p =20% = 0.2

n = 15

a) We have to find P(x >= 6 )

P(x >= 6 ) = 1 - P( x < 6 ) = 1 - P( x <= 5 )

Using Excel function ,   =BINOMDIST( x , n , p , 1 )

P( x <= 5 ) = BINOMDIST( 5 , 15 , 0.2, 1 ) = 0.9389

So, P(x >= 6 ) = 1 - 0.9389 = 0.0611

Probability that 6 or more live in poverty is 0.0611

b) We have to find P( x = 5 )

Using Excel function ,   =BINOMDIST( x , n , p , 0 )

P( x = 5 ) =BINOMDIST( 5 , 15 , 0.2, 0 ) = 0.1032

Probability that 5 live in poverty is 0.1032

c ) Expected value ( mean ) :

Mean = n*p = 0.2*15 = 3