Chapter 8.3, Problem 8.66E

(a)

To determine

To construct a histogram of the probability distribution and how this is different from the one in example.

Explanation of Solution

We note that from Galton’s board:

  $\begin{array}{l}n=6\\ p=0.6\end{array}$

Evaluate at $k=0\text{ to 6}$ :

  $\begin{array}{}$ P(X=0)=C{}_0\times {(0.6)}^0\times {(1-0.6)}^{6-0}=0.004096$$ P(X=1)=C{}_1\times {(0.6)}^1\times {(1-0.6)}^{6-1}=0.0369$$ P(X=2)=C{}_2\times {(0.6)}^2\times {(1-0.6)}^{6-2}=0.13824$$ P(X=3)=C{}_3\times {(0.6)}^3\times {(1-0.6)}^{6-3}=0.2765$$ P(X=4)=C{}_4\times {(0.6)}^4\times {(1-0.6)}^{6-4}=0.3110$$ P(X=5)=C{}_5\times {(0.6)}^5\times {(1-0.6)}^{6-5}=0.1866$$ P(X=6)=C{}_6\times {(0.6)}^6\times {(1-0.6)}^{6-6}=0.0467$\end{array}$

The histogram is as:

  

We note that this distribution is skewed to the left while the graph in the example was symmetric.

(b)

To determine

To calculate how many ball would you expect to land Slot D and explain.

Answer to Problem 8.66E

  $0.27648$ .

Explanation of Solution

The ball ends up in Slot D, if the ball moved to the right of the peg three times and the ball moved to the left of the peg three times.

Evaluate at $k=3$ :

  $\begin{array}{l}$ P(X=3)=C{}_3\times {(0.6)}^3\times {(1-0.6)}^{6-3}$=20\times {0.6}^3\times {0.4}^4\\ =0.27648\end{array}$