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Given in the question that, The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $\$4.59$and a standard deviation of$\$0.10$. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the $16$gas stations.

The cost of unleaded gasoline in the Bay Area, according to the data, has an unknown distribution with a mean of $4.59$and a standard deviation of $0.10.$There are$16$people in the sample.

Assume that $X$is the cost of unleaded gasoline in the San Francisco Bay Area.

$\stackrel{¯}{X}$represents the average cost of fuel for the 16 gas stations.

The Central Limit Theorem states that

According to the Central limit theorem,

$\stackrel{¯}{X}−N\left({\mathrm{μ}}_x,\frac{{\mathrm{σ}}_x}{\sqrt{n}}\right)$

$\stackrel{¯}{X}−N\left(4.59,\frac{0.10}{\sqrt{16}}\right)$

$\stackrel{¯}{X}~N(4.59,0.025)$

The required probability is $p\left(\stackrel{¯}{x}>4.96\right)$

To see the probability that the average price for $16$gas stations is over $\$4.69$, use the $Ti-83$calculator. For this, click on $2^{\text{nd}}$, then DISTR, then scroll down to the normcdf option and fill in the provided information. After that, press the calculator's ENTER key to get the desired result. The following is a screenshot:

Therefore, the probability that average price for $16$gas stations is over $\$4.69$is:

$P(\stackrel{¯}{X}>\mathrm{\$}4.69)=1−P(\stackrel{¯}{X}<\mathrm{\$}4.69)$

$=1−0.99997$

$=0.00003$

Thus, probability is almost$0$. So, the correct option is 'a'