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

We are given that the heights of male statistics students, when wearing shoes, follow a Normal distribution with a mean of 182 cm and a standard deviation of 7.2 cm. The door height is 191 cm, and we need to find the probability that none of the 16 males are taller than this door height, i.e., they don't need to bend down.

### Step 1: Standardize the height
We need to find the probability that a randomly chosen student has a height less than or equal to 191 cm. Let \( X \) be the height of a student, then \( X \sim N(182, 7.2^2) \).

We standardize \( X = 191 \) using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma} = \frac{191 - 182}{7.2} = \frac{9}{7.2} \approx 1.25
\]

### Step 2: Find the cumulative probability
We now need to find \( P(Z \leq 1.25) \) using the standard normal distribution table.

From standard normal distribution tables, the cumulative probability for \( Z = 1.25 \) is approximately 0.8944. This means that for any individual student, the probability of not having to bend down is 0.8944.

### Step 3: Probability that all 16 students don't need to bend down
Since the heights of the 16 students are independent, the probability that none of them need to bend down is the product of each individual probability:
\[
P(\text{none bend down}) = (0.8944)^{16}
\]

Let's calculate this.

\[
P(\text{none bend down}) = 0.8944^{16}
\]

I will compute this value.The probability that none of the 16 students have to bend down when entering the room is approximately \( 0.168 \) or 16.8%.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the probability change if the door height was increased by 5 cm?
2. What is the probability that at least one student will need to bend down?
3. How would the result change if the standard deviation of the students' heights increased to 8 cm?
4. What is the expected number of students who would have to bend down given this setup?
5. How would this analysis differ if the distribution was skewed instead of normal?

**Tip:** When solving probability problems involving multiple independent events, always check if the events truly are independent before multiplying 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 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...

Explore how to calculate the probability that none of the 16 male statistics students, whose heights follow a normal distribution with a mean of 182 cm and a standard deviation of 7.2 cm, will need to bend under a door height of 191 cm. Learn about Z-scores and the use of standard normal distribution tables.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation