The problem describes a normal distribution for the time it takes to find a parking space at 9 a.m., with the following parameters:

• Mean ($\mu$): 5 minutes
• Standard deviation ($\sigma$): 2 minutes

We'll address the two parts of the problem step by step.

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1. Find the probability that it takes at least eight minutes to find a parking space.

We need to find the probability that $X \geq 8$, where $X$ is the time to find a parking space.

This can be expressed as:

$P(X \geq 8) = 1 - P(X < 8)$

We need to standardize the normal variable by converting it into a standard normal variable (z-score) using the formula:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X = 8$ minutes (our target)
• $\mu = 5$ minutes
• $\sigma = 2$ minutes

Now, let's compute the z-score for $X = 8$:

$Z = \frac{8 - 5}{2} = \frac{3}{2} = 1.5$

Next, we look up the probability corresponding to $Z = 1.5$ using a standard normal table (or a calculator).

The cumulative probability for $Z = 1.5$ is approximately:

$P(Z < 1.5) = 0.9332$

Thus:

$P(X \geq 8) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668$

So, the probability that it takes at least eight minutes to find a parking space is approximately 0.0668 (or 6.68%).

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2. Seventy percent of the time, it takes more than how many minutes to find a parking space?

Here, we are looking for the time $X$ such that 70% of the time, it takes more than that time to find a parking space. This implies we are searching for the 30th percentile (since 70% of the time is more than this value).

Let $X$ be the value such that:

$P(X > k) = 0.70 \implies P(X < k) = 0.30$

Using the z-score formula, we can find the corresponding z-score for which 30% of the data lies to the left. From standard normal distribution tables (or a calculator), the z-score for the 30th percentile is approximately:

$Z = -0.524$

Now, we use the z-score formula to solve for $k$ (the time):

$Z = \frac{k - \mu}{\sigma}$

Substitute the values of $Z$, $\mu$, and $\sigma$:

$-0.524 = \frac{k - 5}{2}$

Now, solve for $k$:

$k - 5 = -0.524 \times 2$ $k - 5 = -1.048$ $k = 5 - 1.048 = 3.952$

Thus, 30% of the time, it takes less than 3.95 minutes to find a parking space. Therefore, 70% of the time, it takes more than 3.95 minutes to find a parking space.

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Summary of Results:

1. The probability that it takes at least 8 minutes to find a parking space is approximately 0.0668 (6.68%).
2. Seventy percent of the time, it takes more than 3.95 minutes to find a parking space.

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Do you want more details or have any questions about these calculations?

5 Related Questions:

1. How would the probabilities change if the standard deviation increased to 3 minutes?
2. What is the probability that it takes between 4 and 6 minutes to find a parking space?
3. How would you find the median time to find a parking space using this distribution?
4. If you needed to find a space 95% of the time within a certain number of minutes, how would you calculate that value?
5. What would the probability be if you were to find a parking space in less than 2 minutes?

Tip:

When solving normal distribution problems, always convert to z-scores first to make use of standard tables or software tools for accurate probabilities!