91Ó°ÊÓ

The number of randomly selected cars$n=100$ and probability of success for a white car is $p=0.23$.

Then,

$$\begin{gathered} q=1-p \\ =1-0.23 \\ =0.77 \end{gathered}$$

From the given information,

$$\begin{gathered} np=100×0.23 \\ =23 \\ >5 \end{gathered}$$

$$\begin{gathered} nq=100×0.77 \\ =77 \\ >5 \end{gathered}$$

Here both and are greater than 5. Hence the binomial probability distribution can be approximated to normal distribution.

The mean value is,

$$\begin{gathered} \mathrm{μ}=np \\ =100×0.23 \\ =23 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \mathrm{σ}=\sqrt{npq} \\ =\sqrt{100×0.23×0.77} \\ =4.21 \end{gathered}$$

The probability that fewer than n outcomes appear is expressed as,

$P\left(X<n\right)=P\left(X<n-0.5\right)$

In this case, define random variable X for the number of white cars.

The probability that fewer than 20 cars are white is expressed as,

$$\begin{gathered} P\left(X<20\right)=P\left(X<20-0.5\right) \\ =P\left(X<19.5\right) \end{gathered}$$

The continuity correction is $\mathbf{P}\left(\mathbf{X}\mathbf{<}\mathbf{19}\mathbf{.}\mathbf{5}\right)$.

Obtain z-score using$x=19.5$, $\mathrm{μ}=23,\mathrm{σ}=4.21$as follows:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{19.5-23}{4.21} \\ =-0.83 \end{gathered}$$

The z-score is -0.83

Using the standard normal table, the corresponds to a cumulative right area of 0.2033.

If $P\left(Xorfewer\right)â\copyright ½0.05$then it is significantly low.

In this case, 0.2033 is greater than 0.05. Hence 20 white cars are not significantly low.

Thus, 20 is not significantly a low number of white cars.