Answer
Given below.
Work Step by Step
(a)
The discrete probability distribution.
The discrete probability distribution is obtained as
\[x\] Frequency Probability
4 18 0.1978
5 18 0.1978
6 20 0.2198
7 35 0.3846
The probability of each value of the random variable can be calculated by using the formula:
\[P\left( {{x}_{i}} \right)=\frac{{{f}_{i}}}{N}\]
Here, \[{{f}_{i}}\]is the frequency of each value, \[N\]is the total frequency, and \[P\left( {{x}_{i}} \right)\]is the probability of each value of the random variable, where \[{{x}_{i}}\] is the value of the random variable.
The total frequency is calculated as
\[\begin{align}
& N=\sum{{{f}_{i}}} \\
& =18+18+20+35 \\
& =91
\end{align}\]
(b)
The graph is shown below.
The above graph shows the probability of each value of the random variable. The horizontal axis shows the number of games played in each World Series and the vertical axis shows the probability. The origin is at the point\[\left( 0,0 \right)\]
The probability is plotted corresponding to each value, and a perpendicular line is drawn from the point to the horizontal axis.
(c)
The mean of the random variable \[X\]is 5.8 games.
The number of people waiting in a queue during lunchtime is the random variable in the provided probability distribution, which is represented by\[X\]
The mean of the random variable \[X\]is calculated by adding the multiplied values of each random variable and their corresponding probability.
The required mean is calculated as
\[\begin{align}
& {{\mu }_{X}}=\sum{\left( x\cdot P\left( x \right) \right)} \\
& =\left( 4\times 0.1978 \right)+\left( 5\times 0.1978 \right)+\left( 6\times 0.2198 \right)+\left( 7\times 0.3846 \right) \\
& =5.7912 \\
& \approx 5.8
\end{align}\]
Therefore, the mean of the random variable \[X\]is 5.8 people.
The mean value of the random variable \[X\]can be interpreted as if the data are collected for a long time,then the expected number of games played in each year is 5.8games.
(d)
The standard deviation of the random variable \[X\] is obtained as 1.2games.
The standard deviation of the random variable \[X\]can be calculated using the formula:
\[\begin{align}
& {{\sigma }_{X}}=\sqrt{\sum{\left( {{\left( x-{{\mu }_{X}} \right)}^{2}}\cdot P\left( x \right) \right)}} \\
& =\sqrt{\sum{\left( {{x}^{2}}\cdot P\left( x \right) \right)-\mu _{X}^{2}}}
\end{align}\]
where\[{{\sigma }_{X}}\]is the standard deviation of \[X\], \[{{\mu }_{X}}\]is the mean of\[X\], and \[P\left( x \right)\]is the probability of each value of the random variable.
The required standard deviation can be calculated as
\[\begin{align}
& {{\sigma }_{X}}=\sqrt{\sum{\left( {{x}^{2}}\cdot P\left( x \right) \right)-\mu _{X}^{2}}} \\
& =\sqrt{\left( \left( {{4}^{2}}\times 0.1978 \right)+\left( {{5}^{2}}\times 0.1978 \right)+\left( {{6}^{2}}\times 0.2198 \right)+\left( {{7}^{2}}\times 0.3846 \right) \right)-{{5.8}^{2}}} \\
& =1.2
\end{align}\]
