NHTSA crash tests. Refer to the NHTSA crash tests of new car models, Exercise 4.3 (p. 217). A summary of the driver-side star ratings for the 98 cars in the file is reproduced in the accompanying Minitab printout. Assume that one of the 98 cars is selected randomly and let x equal the number of stars in the car’s driver-side star rating.

1. Use the information in the printout to find the probability distribution for x.
2. Find$\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x=5}}\mathbf{\right)}$.
3. Find$\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{\le }\mathbf{2}\mathbf{)}$.
4. Find $\mathbf{\mu }\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and practically interpret the result.

The formula for calculating the probability distribution isshown below:

$\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\text{=}}\frac{\mathbf{\text{Probability}}}{\mathbf{\text{100}}}$

Here role="math" localid="1653643443072" $\mathbf{X}$,it represents the stars.

The calculation $\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$is shown below:

Therefore, the probability distribution of the associated number of stars is 0.0408, 0.1735, 0.6020 and 0.1837.

The formula for calculating the $\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{=5}\mathbf{)}$is shown below:

$\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{=}\mathbf{X}\mathbf{)}\mathbf{=}\frac{\mathbf{Probability}}{\mathbf{100}}$

Here, as$\mathbf{X}$ it is equal to 5, the associated probability in percentage, 18.37, must be considered.

The calculation $\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x=5}}\mathbf{\right)}$is shown below:

$$\begin{gathered} \mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x=5}}\mathbf{\right)}\mathbf{\text{=}}\frac{\mathbf{\text{18.37}}}{\mathbf{\text{100}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.1837}} \end{gathered}$$

Therefore, the $\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x=5}}\mathbf{\right)}$ is 0.1837.

The formula for calculating the $\mathbf{\text{P}}\mathbf{(}\mathbf{\text{x}}\mathbf{\le }\mathbf{2}\mathbf{)}$is shown below:

$\mathbf{\text{P}}\mathbf{(}\mathbf{\text{x}}\mathbf{\le }\mathbf{\text{2}}\mathbf{)}\mathbf{\text{=P}}\mathbf{\left(}\mathbf{\text{x=0}}\mathbf{\right)}\mathbf{\text{+P}}\mathbf{\left(}\mathbf{\text{x=1}}\mathbf{\right)}\mathbf{\text{+P}}\mathbf{\left(}\mathbf{\text{x=2}}\mathbf{\right)}$

Here the summation of the probabilities from$\mathbf{X}\mathbf{=0}$ to$\mathbf{\text{X=2}}$ must be done.

The calculation $\mathbf{\text{P}}\mathbf{(}\mathbf{\text{x}}\mathbf{\le }\mathbf{2}\mathbf{)}$is shown below:

$$\begin{gathered} \mathbf{\text{P}}\mathbf{(}\mathbf{\text{x}}\mathbf{\le }\mathbf{\text{2}}\mathbf{)}\mathbf{\text{=}}\frac{\mathbf{\text{0}}}{\mathbf{\text{100}}}\mathbf{\text{+}}\frac{\mathbf{\text{0}}}{\mathbf{\text{100}}}\mathbf{\text{+}}\frac{\mathbf{\text{4.08}}}{\mathbf{\text{100}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\frac{\mathbf{\text{4.08}}}{\mathbf{\text{100}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.0408}} \end{gathered}$$

Therefore, the $\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{\le }\mathbf{2}\mathbf{)}$ is 0.0408.

$$\begin{gathered} \mathbf{\text{μ=E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=E}}\mathbf{\left[}\mathbf{\text{xp}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=2×0.0408+3×0.1735+4×0.6020+5×0.1837}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=3.9286}} \end{gathered}$$

Therefore, the $\mathbf{\text{μ=E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$ is 3.9286.

The value$\mathbf{\text{μ=E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$ is found to be 3.9286, which is the average. It indicates that whenever a car is selected at random, there will be a high probability for it to possess 3.9286 stars.