The heights of male statistics students when wearing shoes can be described by a Normal distribution with mean 182 cm and standard deviation 7.2 cm. Suppose 16 males arrive independently to a workshop where the height of the door is 191 cm. The probability that none of the students have to bend down when entering the room is

The problem requires calculating the probability that none of the 16 male students have to bend down when entering a room with a door height of 191 cm, given that their heights (while wearing shoes) follow a Normal distribution with a mean of 182 cm and a standard deviation of 7.2 cm.

### Step-by-step Solution:

1. **Normal distribution of heights**:
   - Mean (\(\mu\)): 182 cm
   - Standard deviation (\(\sigma\)): 7.2 cm
   - Door height: 191 cm

2. **Find the z-score for 191 cm**:
   The z-score allows us to standardize the height value and calculate the probability using the standard Normal distribution table.
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   Where:
   - \(x = 191\) cm (the height of the door)
   - \(\mu = 182\) cm (the mean height)
   - \(\sigma = 7.2\) cm (the standard deviation)

   Substituting the values:
   \[
   z = \frac{191 - 182}{7.2} = \frac{9}{7.2} \approx 1.25
   \]

3. **Find the probability for a z-score of 1.25**:
   Using standard Normal distribution tables or a calculator, the probability that a student’s height is less than 191 cm corresponds to the cumulative probability for \(z = 1.25\).
   
   \[
   P(Z < 1.25) \approx 0.8944
   \]

   This means that approximately 89.44% of male statistics students will not have to bend down when entering the room.

4. **Calculate the probability that none of the 16 students have to bend down**:
   Each student independently has an 89.44% chance of not needing to bend down. For all 16 students, we calculate the probability that all of them will not need to bend down by raising the individual probability to the 16th power:
   
   \[
   P(\text{none bend}) = (0.8944)^{16} \approx 0.201
   \]

### Final Answer:
The probability that none of the 16 students will have to bend down when entering the room is approximately **0.201** or **20.1%**.

Do you want further details or have any questions?

Here are 5 related questions to deepen your understanding:
1. What is the probability that at least one student will have to bend down?
2. How would the probability change if the standard deviation were larger?
3. What is the probability that exactly 5 students will have to bend down?
4. How does the probability change if the mean height is 180 cm?
5. How would the problem change if the number of students were increased to 25?

**Tip:** Z-scores are a useful way to standardize data and compare different datasets or distributions.

Calculate the probability that none of the 16 male students, whose heights follow a Normal distribution (mean 182 cm, standard deviation 7.2 cm), have to bend down when entering a room with a door height of 191 cm. Learn how to apply the z-score and properties of the Normal distribution to solve this problem. Suitable for undergraduate statistics students.

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