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 20 males arrive independently to a workshop where the height of the door is 193 cm. The probability that none of the students have to bend down when entering the room is



To solve this problem, we need to determine the probability that none of the 20 male students will have to bend down when entering a room with a door height of 193 cm.

### Step 1: Determine the probability that a single student does not have to bend down.

Let \( X \) be the height of a male statistics student in shoes, which follows a Normal distribution with mean \( \mu = 182 \) cm and standard deviation \( \sigma = 7.2 \) cm. We want to find the probability that a student's height \( X \) is less than or equal to 193 cm.

This can be calculated as:

\[
P(X \leq 193) = P\left(Z \leq \frac{193 - 182}{7.2}\right)
\]

Where \( Z \) is the standard normal variable.

\[
Z = \frac{193 - 182}{7.2} = \frac{11}{7.2} \approx 1.528
\]

Now, we look up the value of \( Z = 1.528 \) in the standard normal distribution table or use a calculator:

\[
P(Z \leq 1.528) \approx 0.937
\]

So, the probability that a single student does not have to bend down is approximately 0.937.

### Step 2: Determine the probability that none of the 20 students have to bend down.

The probability that all 20 students do not have to bend down is the product of the individual probabilities, assuming independence:

\[
P(\text{None bend down}) = 0.937^{20}
\]

Let's calculate \( 0.937^{20} \):

\[
0.937^{20} \approx 0.375
\]

### Final Answer:
The probability that none of the 20 students have to bend down when entering the room is approximately **0.375** or **37.5%**.

Would you like further details on this solution or have any other questions?

Here are 5 related questions to expand your understanding:
1. What would be the probability that exactly one student has to bend down?
2. How would the probability change if the door height were reduced to 190 cm?
3. How would the problem change if the standard deviation of heights was smaller, say 5 cm?
4. What is the probability that at least one student has to bend down?
5. How would you solve this problem if the number of students was increased to 30?

**Tip:** When dealing with normal distributions, using the Z-score helps standardize values, making it easier to calculate probabilities.

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 20 males arrive independently to a workshop where the height of the door is 193 cm. The probability that none of the students have to bend down when entering the room is

Explore how to calculate the probability that none of 20 male students, with heights normally distributed (mean 182 cm, standard deviation 7.2 cm), have to bend when entering a room with a 193 cm door. Includes a detailed explanation using Z-scores and probability theory.

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