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castortr0y [4]
1 year ago
7

Mrs. Smith gave two classes the same quiz. She randomly selected 5 students from each class to compare the classes. The scores o

f the five students are shown below. Class A: 80, 82, 74, 92, 94 Class B: 82, 80, 88, 86, 82
What are the mean and range for each class ?
Class A- mean = 84.4 range = 14
Class B - mean = 83.6. range = 8

Using the information found in the question above which class mrs smith report had the more consistent performance on the unit test ?, explain what this means in terms of the situation
Mathematics
1 answer:
Tasya [4]1 year ago
4 0

The mean and range for class A are 84.4 and 20 while the mean and range for class B are 83.6 and 8. Class B is more consistant then class A.

<h3>How to calculate the mean?</h3>

We can see from the data that the consistency of class B is more becuase the repeatation of the same score is more in class B then class A  and the range of class B is also less then the class A.

It should be noted that mean simply means average. It's the addition of the numbers divided by the total numbers.

In this case, the computation will be:

Class A: Mean = 422/5= 84.4.

Range: 94-74 = 20

Class B: Mean = 418/5 = 83.6.

Range= 88-80 = 8

Learn more about mean on:

brainly.com/question/1136789

#SPJ1

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3 years ago
Read 2 more answers
Need help pls<br>thank you in advance ​
ira [324]

Answer:

<h3><u>Mean</u></h3>

<u />

\textsf{Mean}\:\overline{X}=\sf \dfrac{\textsf{sum of all the data values}}{\textsf{total number of data values}}

\implies \sf Mean\:(Nilo)=\dfrac{5+6+14+15}{4}=\dfrac{40}{4}=10

\implies \sf Mean\:(Lisa)=\dfrac{8+9+11+12}{4}=\dfrac{40}{4}=10

<h3><u>Standard Deviation</u></h3>

\displaystyle \textsf{Standard Deviation }s=\sqrt{\dfrac{\sum X^2-\dfrac{(\sum X)^2}{n}}{n-1}}

\begin{aligned}\displaystyle \textsf{Standard Deviation (Nilo)} & =\sqrt{\dfrac{(5^2+6^2+14^2+15^2)-\dfrac{(5+6+14+15)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{482-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{82}{3}}\\\\& = 5.23\end{aligned}

\begin{aligned}\displaystyle \textsf{Standard Deviation (Lisa)} & =\sqrt{\dfrac{(8^2+9^2+11^2+12^2)-\dfrac{(8+9+11+12)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{410-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{10}{3}}\\\\& = 1.83\end{aligned}

<h3><u>Summary</u></h3>

Nilo has a mean score of 10 and a standard deviation of 5.23.

Lisa has a mean score of 10 and a standard deviation of 1.83.

The <u>mean</u> scores are the <u>same</u>.

Nilo's standard deviation is higher than Lisa's.  Therefore, Nilo's test scores are more <u>spread out</u> that Lisa's, which means Lisa's test scores are more <u>consistent</u>.

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