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 $P(3) = 4.7*10^{-5}$

Step-by-step 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

       $\sigma = \sqrt{ npq}$

substituting values

      $\sigma = \sqrt{18 * 0.64 * 0.36}$

      $\sigma = 2.036$

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

    $\left n} \atop \right. C_3 = \frac{n! }{[n-3] ! 3!}$

substituting values

    $\left n} \atop \right. C_3 = \frac{18! }{[15] ! 3!}$

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

   $\left n} \atop \right. C_3 = \frac{18 * 17 * 16 }{ (3 *2 *1 )}$

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

So

    $P(3) = 816 * (0.64 )^3 * (0.36 )^{18-3}$

    $P(3) = 4.7*10^{-5}$