2. Use the random variable notation to express symbolically each of the following:

X<0.9
An event in which the z-score of a randomly selected student is less than 0.9.

P(X<-1.3)
The probability of an event in which the z-score of a randomly selected student is less than -1.3.

P(X<1.6)=0.94
The probability of an event in which the z-score of a randomly selected student is less than 1.6 is equal to 0.94.

3. Use technology to find the probability that the z-score of a randomly selected student is:

i. less than 0.9:

(round the answer to 2 decimal places)

ii. more than -1.3:

(round the answer to 2 decimal places)

iii. between -2.2 and 1.7:

(round the answer to 2 decimal places)

iv. more than 9.5:

(round the answer to 2 decimal places)

v. less than 12.2:

(round the answer to 2 decimal places)

4. Find the 40-th percentile:

=
0.40
(round the answer to 2 decimal places)

### 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:

\[ P(X < 0.9) \approx 0.82 \]

---

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:

\[ P(X > -1.3) \approx 0.90 \]

---

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:

\[ P(-2.2 < X < 1.7) \approx 0.94 \]

---

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:

\[ P(X > 9.5) \approx 0.0000 \]

---

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:

\[ P(X < 12.2) \approx 1.0000 \]

---

### 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:

\[ X_p \approx -0.25 \]

---

### 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.

2. Use the random variable notation to express symbolically each of the following: X<0.9 An event in which the z-score of a randomly selected student is less than 0.9. P(X<-1.3) The probability of an event in which the z-score of a randomly selected student is less than -1.3. P(X<1.6)=0.94 The probability of an event in which the z-score of a randomly selected student is less than 1.6 is equal to 0.94. 3. Use technology to find the probability that the z-score of a randomly selected student is: i. less than 0.9: (round the answer to 2 decimal places) ii. more than -1.3: (round the answer to 2 decimal places) iii. between -2.2 and 1.7: (round the answer to 2 decimal places) iv. more than 9.5: (round the answer to 2 decimal places) v. less than 12.2: (round the answer to 2 decimal places) 4. Find the 40-th percentile: = 0.40 (round the answer to 2 decimal places)

Learn how to calculate probabilities for z-scores and find the 40th percentile using the standard normal distribution. This problem involves z-scores, cumulative probabilities, and the use of technology to compute precise values. Suitable for Grades 11-12 or introductory college students.

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