2. Expressing the given events symbolically using random variable notation:

• $X < 0.9$: This represents an event where the z-score of a randomly selected student is less than 0.9.

• $P(X < -1.3)$: This expresses the probability that the z-score of a randomly selected student is less than -1.3.

• $P(X < 1.6) = 0.94$: This indicates that the probability of the z-score of a randomly selected student being less than 1.6 is 0.94.

3. Finding probabilities using the standard normal distribution:

i. $P(X < 0.9)$:

Using technology (standard normal distribution table or calculator), the probability of the z-score being less than 0.9 can be calculated as:

$P(X < 0.9) \approx 0.8159$

Rounded to two decimal places:

$P(X < 0.9) \approx 0.82$

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ii. $P(X > -1.3)$:

First, find $P(X < -1.3)$, which is the cumulative probability:

$P(X < -1.3) \approx 0.0968$

So, the probability of being greater than -1.3 is:

$P(X > -1.3) = 1 - P(X < -1.3) = 1 - 0.0968 = 0.9032$

Rounded to two decimal places:

$P(X > -1.3) \approx 0.90$

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iii. $P(-2.2 < X < 1.7)$:

First, find the cumulative probabilities for $X < 1.7$ and $X < -2.2$:

• $P(X < 1.7) \approx 0.9554$
• $P(X < -2.2) \approx 0.0139$

Now, the probability between -2.2 and 1.7 is:

$P(-2.2 < X < 1.7) = P(X < 1.7) - P(X < -2.2) = 0.9554 - 0.0139 = 0.9415$

Rounded to two decimal places:

$P(-2.2 < X < 1.7) \approx 0.94$

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iv. $P(X > 9.5)$:

Since a z-score of 9.5 is extremely large, the probability that $X > 9.5$ is very close to 0. Using standard normal distribution properties:

$P(X > 9.5) \approx 0.0000$

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v. $P(X < 12.2)$:

A z-score of 12.2 is also extremely large, so the probability that $X < 12.2$ is approximately 1:

$P(X < 12.2) \approx 1.0000$

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4. Finding the 40-th percentile:

The 40-th percentile corresponds to the value of $X$ where $P(X < X_p) = 0.40$. Using technology to find this value:

The z-score for the 40-th percentile is approximately:

$X_p \approx -0.25$

Rounded to two decimal places:

$X_p \approx -0.25$

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Follow-up Questions:

1. How do you interpret the cumulative probability for a standard normal distribution?
2. What is the significance of a z-score being more than 9.5 or less than -9.5?
3. How would you calculate the 60-th percentile for a normal distribution?
4. How would the probabilities change if the distribution were skewed instead of normal?
5. Why is it important to round probabilities in practical applications?

Tip: The z-score is a way of standardizing values from different distributions so they can be compared on the same scale. It's especially useful in understanding relative positions in a normal distribution.