Chapter 11.9, Problem 30PPSE

Solution

To explain: A situation that suits the probability distribution graph given.

The probability that a student misses a class on a day is 0.3 and the distribution is about the probability of a student missing the class in 3 days of school.

Given:

A graph of a probability distribution as shown below

  

The total number of trails $n=3$

The probability of success or student miss a class $p=0.3$

The probability of failure or student does not miss class $q=0.7$

Concept used:

The probability distribution function for a binomial random variable X with parameters n, and p is

  $P(X=x)=n_{C_x}{p}^x{(1-p)}^{n-x}$

Interpretation:From the given information

  $\begin{array}{l}n=3\\ p=0.3\end{array}$

A situation where a student misses class on each day with a probability of 0.3 and the probability distribution for 3 days is being calculated. This is an apt situation for the given probability distribution.

The probabilities are calculated as shown below

  $\begin{array}{c}$ P(X=0)=n_{C_x}{p}^x{(1-p)}^{n-x}$$ =3_{C_0}{0.3}^0{(1-0.3)}^{3-0}$=1\times 1\times 0.343\\ =0.343\end{array}$

  $\begin{array}{c}$ P(X=1)=n_{C_x}{p}^x{(1-p)}^{n-x}$$ =3_{C_1}{0.3}^1{(1-0.3)}^{3-1}$=3\times 0.3\times 0.49\\ =0.441\end{array}$

  $\begin{array}{c}$ P(X=2)=n_{C_x}{p}^x{(1-p)}^{n-x}$$ =3_{C_2}{0.3}^2{(1-0.3)}^{3-2}$=3\times 0.09\times 0.7\\ =0.189\end{array}$

  $\begin{array}{c}$ P(X=3)=n_{C_x}{p}^x{(1-p)}^{n-x}$$ =3_{C_3}{0.3}^3{(1-0.3)}^{3-3}$=1\times 0.027\times 1\\ =0.027\end{array}$

Thus the probabilities with respect to the given graph are calculated and the given situation is an apt for the given probability graph.