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In this exercise, we are dealing with a Gaussian (normal) random variable \( X \) with a mean \( m = 50 \) and standard deviation \( \sigma = 4 \). The task is to compute or approximate the following probabilities:

1. \( P(X > 40) \)
2. \( P(X < 60) \)
3. \( P(X > 90) \)
4. \( P(60 < X < 80) \)
5. \( P(|X - 40| < 10) \)

### Strategy:

We will use the standard normal distribution table (Z-table) to solve this by converting the values of \( X \) into their corresponding \( Z \)-scores using the formula:
\[
Z = \frac{X - m}{\sigma}
\]
where \( m \) is the mean and \( \sigma \) is the standard deviation.

Let me calculate each probability.

### (1) \( P(X > 40) \)
For this, we need to calculate the Z-score for \( X = 40 \):
\[
Z = \frac{40 - 50}{4} = -2.5
\]
We are looking for \( P(X > 40) = P(Z > -2.5) \). Using the Z-table, \( P(Z < -2.5) \approx 0.0062 \). So,
\[
P(X > 40) = 1 - P(Z < -2.5) = 1 - 0.0062 = 0.9938
\]

### (2) \( P(X < 60) \)
For \( X = 60 \), the Z-score is:
\[
Z = \frac{60 - 50}{4} = 2.5
\]
We are looking for \( P(X < 60) = P(Z < 2.5) \). From the Z-table, \( P(Z < 2.5) \approx 0.9938 \), so:
\[
P(X < 60) = 0.9938
\]

### (3) \( P(X > 90) \)
For \( X = 90 \), the Z-score is:
\[
Z = \frac{90 - 50}{4} = 10
\]
A Z-score of 10 is extremely high, and the probability is essentially 0:
\[
P(X > 90) \approx 0
\]

### (4) \( P(60 < X < 80) \)
For \( X = 60 \) and \( X = 80 \), the Z-scores are:
\[
Z_{60} = \frac{60 - 50}{4} = 2.5, \quad Z_{80} = \frac{80 - 50}{4} = 7.5
\]
We already know from the Z-table that \( P(Z < 2.5) \approx 0.9938 \), and since \( P(Z < 7.5) \approx 1 \) (very high), the probability is:
\[
P(60 < X < 80) = P(Z < 7.5) - P(Z < 2.5) = 1 - 0.9938 = 0.0062
\]

### (5) \( P(|X - 40| < 10) \)
This condition means \( 30 < X < 50 \). The Z-scores are:
\[
Z_{30} = \frac{30 - 50}{4} = -5, \quad Z_{50} = \frac{50 - 50}{4} = 0
\]
For \( Z = -5 \), the probability is essentially 0, and for \( Z = 0 \), \( P(Z < 0) = 0.5 \). Therefore:
\[
P(|X - 40| < 10) = P(30 < X < 50) \approx 0.5
\]

Let me know if you would like more details or explanations on any of these. Here are 5 questions related to this:

1. What is the general formula to convert any normal variable into a standard normal variable?
2. How does the Z-score relate to standard deviation?
3. Can you explain why the probability of \( P(X > 90) \) is practically 0?
4. How does the Z-table help in calculating probabilities for normal distributions?
5. What would happen to the probabilities if the standard deviation \( \sigma \) were larger or smaller?

**Tip:** Always sketch the distribution curve when working with normal distributions—it helps visualize the areas corresponding to probabilities!

Soit X une variable aléatoire gaussienne de moyenne m = 50 et d'écart-type σ = 4. Approcher: 1) P(X > 40) 2) P(X < 60) 3) P(X > 90) 4) P(60 < X < 80) 5) P(|X - 40| < 10).

Learn how to calculate probabilities for a Gaussian random variable X with mean 50 and standard deviation 4. This includes P(X > 40), P(X < 60), and other important Gaussian distribution scenarios using Z-scores and normal distribution tables.

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