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Thank you for visiting Data Table 99 1 98 8 98 5 97 7 98 7 98 6 98 98 1 98 99 98 5 99 99 1 96. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

Data Table:

99.1, 98.8, 98.5, 97.7, 98.7, 98.6, 98, 98.1, 98, 99, 98.5, 99, 99.1, 96.9, 97.2, 98.7, 97.9, 97.3, 99.2, 98.3, 97.7, 98.1, 98.3, 97.3, 99.1, 97.7, 99.2, 98.4, 99.4, 97.9, 97.5, 98, 98.6, 99.4, 98.3, 99.1, 98.2, 98.4, 98.3, 98.8, 96.9, 98.3, 99, 98.4, 97.9, 98.3, 98.1, 98.4, 98.3, 98.4, 97.9, 99.1, 97, 97.6, 98.3, 96.6, 97.5, 97.5, 97.7, 98.1, 97.3, 97.7, 97, 97.3, 98.2, 98.7, 99.2, 99, 97.5, 97.7, 97.5, 97.4, 99.8, 98.4, 98, 98.5, 98, 99, 98, 98.5, 98.4, 98.7, 98.4, 97.5, 99.3, 97.8, 98.6, 98.9, 99.3, 98.5, 98.7, 96.6, 97.6, 98.3, 98.6, 97.9, 97.9, 99, 98.1, 98.1, 98.6, 97.7, 98.5, 97.6, 97.9, 96.7, 98.5, 98.7, 98.6

Answer :

Therefore, the required range will be 3.20.

Therefore, the required variance will be 1.12.

Therefore, the required standard deviation will be 1.06.

To find the range, variance, and standard deviation of the given body temperature data, follow these steps:

Range: The range is the difference between the maximum and minimum values in the dataset.

Range = Maximum Value - Minimum Value

Variance: The variance measures how much the data points deviate from the mean. It's calculated as the average of the squared differences from the mean.

Variance = [tex]\[\frac{\sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}\][/tex]

Standard Deviation: The standard deviation is the square root of the variance and provides a measure of the dispersion of the data.

Standard Deviation = [tex]\sqrt{Variance}[/tex]

We'll calculate the results step by step:

Range:

Range = Maximum Value - Minimum Value

Range = 99.8 - 96.6 = 3.2

Variance:

Calculate the mean([tex]\overline{x}[/tex]):

[tex]$\overline{x}=\frac{\sum{{{x}_{i}}}}{n}=\frac{3292.3}{100}=32.923$[/tex]

Calculate the sum of squared differences from the mean:

[tex]$\sum{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}=111.6521$[/tex]

Variance = [tex]$\frac{\sum{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}}{n}=\frac{111.6521}{100}=1.1165$[/tex]

Standard Deviation:

Standard Deviation = [tex]\sqrt{Variance}[/tex] = [tex]\sqrt{1.1165}[/tex] ≈ 1.0567

Learn more about range at:

https://brainly.com/question/35386870

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The question is-

Find the Range, variance, and standard deviation of the following body temperatures, in degrees Fahrenheit, taken at 12:00AM (Round to two decimal places as needed).

Given the dataset: 99.1 98.8 98.5 97.7 98.7 98.6 98 98.1 98 99 98.5 99 99.1 96.9 97.2 98.7 97.9 97.3 99.2 98.3 97.7 98.1 98.3 97.3 99.1 97.7 99.2 98.4 99.4 97.9 97.5 98 98.6 99.4 98.3 99.1 98.2 98.4 98.3 98.8 96.9 98.3 99 98.4 97.9 98.3 98.1 98.4 98.3 98.4 97.9 99.1 97 97.6 98.3 96.6 97.5 97.5 97.7 98.1 97.3 97.7 97 97.3 98.2 98.7 99.2 99 97.5 97.7 97.5 97.4 99.8 98.4 98. 98.5 98 99 98 98.5 98.4 98.7 98.4 97.5 99.3 97.8 98.6 98.9 99.3 98.5 98.7 96.6 97.6 98.3 98.6 97.9 97.9 99 98.1 98.1 98.6 97.7 98.5 97.6 97.9 96.7 98.5 98.7 98.6

Thank you for reading the article Data Table 99 1 98 8 98 5 97 7 98 7 98 6 98 98 1 98 99 98 5 99 99 1 96. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!

Rewritten by : Jeany