ACTIVITY NO. 10
The mean height of the learners in Grade 1 Newton is
125 cm and the standard deviation is 3.5 cm . Assuming the
heights are normally, what percent of the height is:
a. Greater than 135 cm ?
b. Less than 128 cm ?
c. Between 115 and 130 cm ?

Question

UpStudy AI · Accepted Answer

To determine the percentages of learners' heights in Grade 1 Newton that fall within specific ranges, we'll use the properties of the normal distribution. Here's a step-by-step solution for each part of the problem.

### Given:
- **Mean (μ)** = 125 cm
- **Standard Deviation (σ)** = 3.5 cm
- **Distribution**: Normal

We'll use the **Z-score formula** to standardize the heights:
$$ Z = \frac{X - \mu}{\sigma} $$

### a. What percent of the heights are **greater than 135 cm**?

1. **Calculate the Z-score for 135 cm:**
   $$
   Z = \frac{135 - 125}{3.5} = \frac{10}{3.5} \approx 2.86
   $$

2. **Find the probability** corresponding to $$ Z = 2.86 $$:
   - Using standard normal distribution tables or a calculator, $$ P(Z < 2.86) \approx 0.9979 $$.
   
3. **Calculate the probability of heights greater than 135 cm:**
   $$
   P(X > 135) = 1 - P(Z < 2.86) = 1 - 0.9979 = 0.0021 \text{ or } 0.21\%
   $$

### **Answer a:** Approximately **0.21%** of the learners are taller than 135 cm.

---

### b. What percent of the heights are **less than 128 cm**?

1. **Calculate the Z-score for 128 cm:**
   $$
   Z = \frac{128 - 125}{3.5} = \frac{3}{3.5} \approx 0.86
   $$

2. **Find the probability** corresponding to $$ Z = 0.86 $$:
   - Using standard normal distribution tables or a calculator, $$ P(Z < 0.86) \approx 0.8051 $$.

### **Answer b:** Approximately **80.51%** of the learners are shorter than 128 cm.

---

### c. What percent of the heights are **between 115 cm and 130 cm**?

1. **Calculate the Z-scores for 115 cm and 130 cm:**
   - For 115 cm:
     $$
     Z = \frac{115 - 125}{3.5} = \frac{-10}{3.5} \approx -2.86
     $$
   - For 130 cm:
     $$
     Z = \frac{130 - 125}{3.5} = \frac{5}{3.5} \approx 1.43
     $$

2. **Find the probabilities** corresponding to these Z-scores:
   - $$ P(Z < 1.43) \approx 0.9236 $$
   - $$ P(Z < -2.86) \approx 1 - P(Z < 2.86) = 1 - 0.9979 = 0.0021 $$

3. **Calculate the probability of heights between 115 cm and 130 cm:**
   $$
   P(115 < X < 130) = P(Z < 1.43) - P(Z < -2.86) = 0.9236 - 0.0021 = 0.9215 \text{ or } 92.15\%
   $$

### **Answer c:** Approximately **92.15%** of the learners have heights between 115 cm and 130 cm.

---

### **Summary of Answers:**
- **a.** Greater than 135 cm: **0.21%**
- **b.** Less than 128 cm: **80.51%**
- **c.** Between 115 cm and 130 cm: **92.15%**
- **a.** 0.21% of learners are taller than 135 cm. - **b.** 80.51% of learners are shorter than 128 cm. - **c.** 92.15% of learners have heights between 115 cm and 130 cm.

ACTIVITY NO. 10
The mean height of the learners in Grade 1 Newton is
125 cm and the standard deviation is 3.5 cm . Assuming the
heights are normally, what percent of the height is:
a. Greater than 135 cm ?
b. Less than 128 cm ?
c. Between 115 and 130 cm ?

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ACTIVITY NO. 10 The mean height of the learners in Grade 1 Newton is 125 cm and the standard deviation is 3.5 cm . Assuming the heights are normally, what percent of the height is: a. Greater than 135 cm ? b. Less than 128 cm ? c. Between 115 and 130 cm ?

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ACTIVITY NO. 10
The mean height of the learners in Grade 1 Newton is
125 cm and the standard deviation is 3.5 cm . Assuming the
heights are normally, what percent of the height is:
a. Greater than 135 cm ?
b. Less than 128 cm ?
c. Between 115 and 130 cm ?