The height of a tree will be measured using a clinometer. For this device, the measurement error has a normal distribution with mean μ=0.00 meter and standard deviation σ=0.05. If the tree is exactly 30 meters tall, what is the probability that the clinometer will indicate a value less than 29.95 meters?

Oh, the antics of measurement errors! Let's calculate the probability of the clinometer indicating a value less than 29.95 meters, shall we?

Since the measurement error follows a normal distribution, we can utilize some math tricks. We'll use the z-score formula to standardize the values.

First, let's find the z-score of 29.95 meters. The formula for z-score is (x - μ) / σ, where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.

So, the z-score of 29.95 meters would be: (29.95 - 30) / 0.05.

Let me do a little juggling here... *calculating z-score*... Ah, the z-score is -1.

Now, to find the probability, we need to use a z-table. Looking up the z-score of -1 in the table, we find the probability is approximately 0.1587.

So, there's about a 15.87% probability that the clinometer will indicate a value less than 29.95 meters. Keep in mind, though, that this probability is based on the assumption that the measurement error follows a normal distribution.

To find the probability that the clinometer will indicate a value less than 29.95 meters, we need to calculate the z-score and then find the corresponding probability from the standard normal distribution.

The formula to calculate the z-score is:

z = (x - μ) / σ

where:
- x is the value we want to calculate the probability for (29.95 meters),
- μ is the mean of the distribution (0.00 meters),
- σ is the standard deviation of the distribution (0.05 meters).

Calculating the z-score:

z = (29.95 - 0.00) / 0.05
z = 599

Next, we need to find the probability corresponding to a z-score of 599 from the standard normal distribution. However, since the z-score is very large, we can use an approximation called the standard normal distribution table.

Based on the standard normal distribution table, the probability corresponding to a z-score of 599 is essentially 1.0.

Therefore, the probability that the clinometer will indicate a value less than 29.95 meters is approximately equal to 1.0.

To calculate the probability that the clinometer will indicate a value less than 29.95 meters, we need to use the properties of a normal distribution.

A normal distribution is characterized by its mean (μ) and standard deviation (σ). In this case, we are given that the mean (μ) is 0.00 meter and the standard deviation (σ) is 0.05 meter.

To solve this problem, we need to calculate the z-score, which represents the number of standard deviations an observation is from the mean. The formula for the z-score is:

z = (x - μ) / σ

where x is the desired height (29.95 meters), μ is the mean (0.00 meter), and σ is the standard deviation (0.05 meter).

Plugging in the values, we have:

z = (29.95 - 0.00) / 0.05 = 599

Now, we need to find the probability that the z-score is less than 599. We can use a standard normal distribution table or a software program to find the corresponding probability.

Looking up the z-score of 599 in a standard normal distribution table, we find that the probability is extremely close to 1 (0.999999999998), which is approximately equal to 1.

Therefore, the probability that the clinometer will indicate a value less than 29.95 meters is approximately 1.