and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. 61-percentile. This is the temperature reading separating the bottom 61% from the top 39% C Submit Question

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**Title: Understanding Normal Distribution through Thermometer Readings**

**Introduction**

In this lesson, we will explore the concept of normal distribution using a practical example involving thermometer readings at the freezing point. We will calculate the temperature corresponding to the 61st percentile, which effectively separates the bottom 61% of the temperatures from the top 39%.

**Problem Statement**

Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of \(0^\circ C\) and a standard deviation of \(1.00^\circ C\). A single thermometer is randomly selected and tested. We aim to find \( P_{61} \), the 61st percentile. This is the temperature reading separating the bottom 61% from the top 39%.

**Calculation**

To find the 61st percentile in a normally distributed dataset with a mean (\( \mu \)) of \(0^\circ C\) and a standard deviation (\( \sigma \)) of \(1.00^\circ C\), we use the z-score corresponding to the 61st percentile. The z-score can be looked up in standard normal distribution tables or calculated using statistical software.

\[ P_{61} = \mu + z_{0.61} \cdot \sigma \]

**Interactive Component**

There is an interactive feature where students can input their calculated \( P_{61} \) and submit their answers for immediate feedback.

**Interactive Box**

\[ P_{61} = \_\_\_\_\_^\circ C \]

**Submit Question Button**

This section allows students to check their calculations and understand the percentile values in the context of normal distribution.

Let's delve into this straightforward yet essential concept of normal distribution and apply it to scenarios beyond thermometer readings!

**Note:** The actual z-score for the 61st percentile would need to be input manually or obtained using a statistical tool.
Transcribed Image Text:**Title: Understanding Normal Distribution through Thermometer Readings** **Introduction** In this lesson, we will explore the concept of normal distribution using a practical example involving thermometer readings at the freezing point. We will calculate the temperature corresponding to the 61st percentile, which effectively separates the bottom 61% of the temperatures from the top 39%. **Problem Statement** Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of \(0^\circ C\) and a standard deviation of \(1.00^\circ C\). A single thermometer is randomly selected and tested. We aim to find \( P_{61} \), the 61st percentile. This is the temperature reading separating the bottom 61% from the top 39%. **Calculation** To find the 61st percentile in a normally distributed dataset with a mean (\( \mu \)) of \(0^\circ C\) and a standard deviation (\( \sigma \)) of \(1.00^\circ C\), we use the z-score corresponding to the 61st percentile. The z-score can be looked up in standard normal distribution tables or calculated using statistical software. \[ P_{61} = \mu + z_{0.61} \cdot \sigma \] **Interactive Component** There is an interactive feature where students can input their calculated \( P_{61} \) and submit their answers for immediate feedback. **Interactive Box** \[ P_{61} = \_\_\_\_\_^\circ C \] **Submit Question Button** This section allows students to check their calculations and understand the percentile values in the context of normal distribution. Let's delve into this straightforward yet essential concept of normal distribution and apply it to scenarios beyond thermometer readings! **Note:** The actual z-score for the 61st percentile would need to be input manually or obtained using a statistical tool.
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