Question

a ride - sharing company has computed its mean fare to be $37.00, with a standard deviation of $4.10. suppose that the fares are normally distributed. complete the following statements. (a) approximately? of the companys rides have fares between $24.70 and $49.30. (b) approximately 75% 95% 99.7% of the companys rides have fares between $ and $.

Explanation:

Step1: Recall the empirical rule for normal distribution

The empirical rule states that for a normal - distributed data: approximately 68% of the data lies within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations of the mean.
Let $\mu = 37.00$ be the mean and $\sigma=4.10$ be the standard deviation.

Step2: Calculate the number of standard - deviations for part (a)

For $x_1 = 24.70$, the z - score is $z_1=\frac{24.70 - 37.00}{4.10}=\frac{- 12.3}{4.10}=-3$.
For $x_2 = 49.30$, the z - score is $z_2=\frac{49.30 - 37.00}{4.10}=\frac{12.3}{4.10}=3$.
Since the data is within 3 standard deviations of the mean ($z=-3$ to $z = 3$), approximately 99.7% of the company's rides have fares between $24.70$ and $49.30$.

Step3: Find the values for part (b) for 95% of the data

For 95% of the data in a normal distribution, the data lies within 2 standard deviations of the mean.
The lower value $x_{lower}=\mu - 2\sigma=37.00-2\times4.10=37.00 - 8.20 = 28.80$.
The upper value $x_{upper}=\mu + 2\sigma=37.00+2\times4.10=37.00 + 8.20 = 45.20$.

Answer:

(a) 99.7%
(b) 28.80, 45.20