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 ?

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%**