Given that the ride-sharing company has computed its mean fare to be $33.00, with a standard deviation of $4.10, this implies that
[tex]\begin{gathered} \mu=33.00 \\ \sigma=4.10 \end{gathered}[/tex]The z score value is expressed as
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ where \\ x\Rightarrow observed\text{ value} \\ \mu\Rightarrow mean\text{ of the sample} \\ \sigma\Rightarrow standard\text{ deviation of the sample} \end{gathered}[/tex]A) Approximately 68% of the company's rides have fares between . . .
From the normal distribution table,
this implies that the z score value is
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