To solve these questions, we will use the properties of a normal distribution, where the mean, $\mu$, is 107500 vehicles and the standard deviation, $\sigma$, is 8800 vehicles.

Given:

• Mean, $\mu = 107500$
• Standard Deviation, $\sigma = 8800$

(a) Probability that the total vehicle usage is more than 120500 vehicles

To find $P(X > 120500)$, we will calculate the $z$-score for 120500.

The $z$-score formula is: $z = \frac{X - \mu}{\sigma}$ Substitute the values: $z = \frac{120500 - 107500}{8800} = \frac{13000}{8800} \approx 1.477$

Using a standard normal distribution table or calculator, we find the probability for $z > 1.477$.

$P(X > 120500) = 1 - P(Z \leq 1.477) \approx 1 - 0.9306 = 0.0694$

So, the probability that the total vehicle usage will be more than 120500 vehicles is approximately 0.0694 or 6.94%.

(b) Probability that the total vehicle usage will be between 80000 and 101000 vehicles

For $P(80000 < X < 101000)$, we calculate the $z$-scores for both 80000 and 101000.

1. For 80000: $z = \frac{80000 - 107500}{8800} = \frac{-27500}{8800} \approx -3.125$

2. For 101000: $z = \frac{101000 - 107500}{8800} = \frac{-6500}{8800} \approx -0.739$

Using a standard normal distribution table or calculator:

$P(Z < -3.125) \approx 0.0009$ $P(Z < -0.739) \approx 0.2296$

Therefore: $P(80000 < X < 101000) = P(Z < -0.739) - P(Z < -3.125) \approx 0.2296 - 0.0009 = 0.2287$

So, the probability that the total vehicle usage will be between 80000 and 101000 vehicles is approximately 0.2287 or 22.87%.

(c) 91st percentile for total vehicle usage

To find the 91st percentile, we need the $z$-score that corresponds to the cumulative probability of 0.91.

From the standard normal distribution, $z_{0.91} \approx 1.34$.

Now, use the $z$-score formula to find $X$: $X = \mu + z \cdot \sigma = 107500 + 1.34 \cdot 8800 \approx 107500 + 11792 = 119292$

Thus, the 91st percentile is approximately 119292 vehicles.

(d) 9th percentile for total vehicle usage

To find the value that corresponds to the top 9% (91st percentile from below), we use the same approach as in part (c). Since we already calculated it, the answer is 119292 vehicles