Question

A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a sample of 18 adults. a-1. How many of the 18 would we expect to indicate that the Treasury should continue making pennies

Answer 1

Complete Question

A recent CBS News survey reported that 64% of adults felt the U.S. Treasury should continue making pennies. Suppose we select a sample of 18 adults.

a-1. How many of the 18 would we expect to indicate that the Treasury should continue making pennies

a-2) What is the standard deviation?

a-3) What is the likelihood that exactly 3 adults would indicate the Treasury should continue making pennies?

Answer:

a-1
`\= x = 11 .52`

a-2
`\sigma = 2.036`

a-3
`P(3) = 4.7*10^(-5)`

Explanation:

From the question we are told that

The sample size is
`n = 18`

The proportion of adult that felt the U.S. Treasury should continue making pennies is p = 0.64

The proportion of adult that feel otherwise is

`q = 1- p = 1-0.64 = 0.36`

The mean is mathematically evaluated as

`\= x = n * p`

substituting values

`\= x = 18 * 0.64`

`\= x = 11 .52`

The standard deviation is mathematically represented as

`\sigma = √( npq)`

substituting values

`\sigma = √(18 * 0.64 * 0.36)`

`\sigma = 2.036`

The likelihood that 3 adult would indicate the Treasury should continue making pennies is mathematically evaluated as

`P(3) = \left n} \atop \right. C_3 (p)^(3) * (q)^(n-3)`

Now

`\left n} \atop \right. C_3 = (n! )/([n-3] ! 3!)`

substituting values

`\left n} \atop \right. C_3 = (18! )/([15] ! 3!)`

`\left n} \atop \right. C_3 = (18 * 17 * 16 * 15! )/([15] ! (3 *2 *1 ))`

`\left n} \atop \right. C_3 = (18 * 17 * 16 )/( (3 *2 *1 ))`

`\left n} \atop \right. C_3 = 816`

So

`P(3) = 816 * (0.64 )^3 * (0.36 )^(18-3)`

`P(3) = 4.7*10^(-5)`