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 
a-2 
a-3 
Step-by-step explanation:
From the question we are told that
The sample size is 
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

The mean is mathematically evaluated as

substituting values


The standard deviation is mathematically represented as

substituting values


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

Now
![\left n} \atop \right. C_3 = \frac{n! }{[n-3] ! 3!}](https://tex.z-dn.net/?f=%5Cleft%20%20n%7D%20%5Catop%20%20%5Cright.%20C_3%20%3D%20%20%5Cfrac%7Bn%21%20%7D%7B%5Bn-3%5D%20%21%203%21%7D)
substituting values
![\left n} \atop \right. C_3 = \frac{18! }{[15] ! 3!}](https://tex.z-dn.net/?f=%5Cleft%20%20n%7D%20%5Catop%20%20%5Cright.%20C_3%20%3D%20%20%5Cfrac%7B18%21%20%7D%7B%5B15%5D%20%21%203%21%7D)
![\left n} \atop \right. C_3 = \frac{18 * 17 * 16 * 15! }{[15] ! (3 *2 *1 )}](https://tex.z-dn.net/?f=%5Cleft%20%20n%7D%20%5Catop%20%20%5Cright.%20C_3%20%3D%20%20%5Cfrac%7B18%20%2A%2017%20%2A%2016%20%2A%2015%21%20%7D%7B%5B15%5D%20%21%20%283%20%2A2%20%2A1%20%29%7D)

So