Answer :
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 [tex]\= x = 11 .52[/tex]
a-2 [tex]\sigma = 2.036[/tex]
a-3 [tex]P(3) = 4.7*10^{-5}[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 18[/tex]
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
[tex]q = 1- p = 1-0.64 = 0.36[/tex]
The mean is mathematically evaluated as
[tex]\= x = n * p[/tex]
substituting values
[tex]\= x = 18 * 0.64[/tex]
[tex]\= x = 11 .52[/tex]
The standard deviation is mathematically represented as
[tex]\sigma = \sqrt{ npq}[/tex]
substituting values
[tex]\sigma = \sqrt{18 * 0.64 * 0.36}[/tex]
[tex]\sigma = 2.036[/tex]
The likelihood that 3 adult would indicate the Treasury should continue making pennies is mathematically evaluated as
[tex]P(3) = \left n} \atop \right. C_3 (p)^{3} * (q)^{n-3}[/tex]
Now
[tex]\left n} \atop \right. C_3 = \frac{n! }{[n-3] ! 3!}[/tex]
substituting values
[tex]\left n} \atop \right. C_3 = \frac{18! }{[15] ! 3!}[/tex]
[tex]\left n} \atop \right. C_3 = \frac{18 * 17 * 16 * 15! }{[15] ! (3 *2 *1 )}[/tex]
[tex]\left n} \atop \right. C_3 = \frac{18 * 17 * 16 }{ (3 *2 *1 )}[/tex]
[tex]\left n} \atop \right. C_3 = 816[/tex]
So
[tex]P(3) = 816 * (0.64 )^3 * (0.36 )^{18-3}[/tex]
[tex]P(3) = 4.7*10^{-5}[/tex]