Answer:
Explanation:
Firstly,let us calculate the probability of getting a tail
If a coin is flipped, we either get a tail or a head show up
The sum of their probabilities is thus 1
So, to get the probability of getting a tail, we subtract the probability of getting a head from 1
We have the probability of getting a tail as:
[tex]1-0.6\text{ = 0.4}[/tex]a) Probability of getting 4 heads
If we get 4 heads, it means we are getting 8 tails
As a general rule, to get r heads from n throws, we have it computed as:
[tex]^nC_r\text{ }\times p^r\times q^{n-r}[/tex]where p is the probability of getting a head and q is the probability of getting a tail
The probability is thus:
[tex]12C4\times(0.6)^4\times(0.4)^8\text{ = 0.042}[/tex]b) The probability of getting at most 5 heads
What this mean is that we can get 0 head (12 tails), 1 head (11 tails), 2 heads , 3 heads , 4 heads or 5 heads
We have to make computations for this as follows:
[tex]\begin{gathered} (12C0\text{ }\times0.4^{12})\text{ + (12C1 }\times0.6^{}\times0.4^{11})\text{ + (12C2}\times0.6^2\text{ }\times0.4^{10})\text{ + (12C3 }\times0.6^3\times0.4^9)\text{ + (12C}4\times\times \\ \times0.6^4\times0.4^8)\text{ + (12C5 }\times0.6^5\times0.4^7) \\ =0.158 \end{gathered}[/tex]c) The probability of achieving more than 8 heads
That means we can get 9 heads (3 tails) , 10 heads( 2 tails) , 11 heads (1 tail) and 12 heads (0 tail)
Mathematically, we have the computation as follows:
[tex]\begin{gathered} (12C9\text{ }\times0.6^9\times0.4^3)\text{ + (12C10}\times0.6^{10}\times0.4^2)\text{ + (12}C11\times0.6^{11}\times0.4)\text{ + (12C12}\times0.6^{12}\times \\ 0.4^0)\text{ = 0}.225 \end{gathered}[/tex]d) Here, we want to get the mean , SD and variance
We have the mean calculated as follows:
[tex]\text{mean = np }[/tex]n is the number of tosses while p is the probabbility of getting a head
n is 12 flips and p is 0.6
By computation, we have the value calculated as:
[tex]\operatorname{mean}\text{ = 12 }\times\text{ 0.6 = 7.2}[/tex]We can calculate the variance as follows:
[tex]\text{variance = n}\times p\times(1-p)[/tex]The meaning of the variables stay same
Thus, we have::
[tex]\text{variance = 12}\times0.6\times(1-0.6)\text{ = 2.88}[/tex]Standad Deviation is the square root of variance which we can have as follows:
[tex]\begin{gathered} \text{ Standard Deviation = }\sqrt[]{n\times p\times(1-p)} \\ =\text{ }\sqrt[]{2.88}\text{ = 1.697} \end{gathered}[/tex]