We are given the following binomial probability distribution
[tex]p(x)=\binom{5}{x}\cdot049^x\cdot0.51^{5-x}[/tex]Recall that the general form of the binomial probability distribution is given by
[tex]p(x)=\binom{n}{x}\cdot p^x\cdot(1-p)^{n-x}[/tex]Comparing the given binomial probability distribution with the general form,
The probability of success is p = 0.49
The sample size is n = 5
Let us find the probability for x = 0, 1, 2, 3, 4, 5
[tex]\begin{gathered} p(x=0)=\binom{5}{0}\cdot049^0\cdot0.51^{5-0}=0.0345 \\ p(x=1)=\binom{5}{1}\cdot049^1\cdot0.51^{5-1}=0.1658 \\ p(x=2)=\binom{5}{2}\cdot049^2\cdot0.51^{5-2}=0.3185 \\ p(x=3)=\binom{5}{3}\cdot049^3\cdot0.51^{5-3}=0.306 \\ p(x=4)=\binom{5}{4}\cdot049^4\cdot0.51^{5-4}=0.147 \\ p(x=5)=\binom{5}{5}\cdot049^5\cdot0.51^{5-5}=0.0283 \end{gathered}[/tex]Now compare the above probabilities with the given graphs.
Notice that the above probabilities exactly matches with graph A
Therefore, the correct answer is graph A.
(b) Mean and standard deviation:
Recall that the mean of the binomial probability distribution is given by
[tex]\mu=n\cdot p[/tex]Where n is the sample size and p is the probability of success
[tex]\mu=n\cdot p=5\cdot0.49=2.45[/tex]Therefore, the mean is 2.45
Recall that the standard deviation of the binomial probability distribution is given by
[tex]\begin{gathered} \sigma=\sqrt[]{n\cdot p\cdot(1-p)} \\ \sigma=\sqrt[]{5\cdot0.49\cdot(1-0.49)} \\ \sigma=1.12 \end{gathered}[/tex]Therefore, the standard deviation is 1.12
(c) The mean is 2.45 and the standard deviation is 1.12
We need to show two standard deviation interval below and above the mean.
[tex]\begin{gathered} lower=\mu-2\cdot\sigma=2.45-2\cdot1.12=2.45-2.24=0.21 \\ upper=\mu+2\cdot\sigma=2.45+2\cdot1.12=2.45+2.24=4.69 \end{gathered}[/tex]So, the two points are 0.21 and 4.69
As you can see, graph C shows these two points.
