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lukranit [14]
2 years ago
14

Class a had a mean score of 7.8 and a standard deviation of 1 class b had a mean score of 8.1 and a standard deviation of 0.2 wh

ich class scored better on average?
Mathematics
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

Class 2 scored better on average

<h3>How to determine which class scored better on average?</h3>

The averages and the standard deviations are given as:

Class 1

Mean score = 7.8

Standard deviation = 1

Class 2

Mean score = 8.1

Standard deviation = 0.2

As a general rule, the smaller the standard deviation; the better the average

Class 2 has a smaller deviation

Hence, class 2 scored better on average

Read more about standard deviation at:

brainly.com/question/28061243

#SPJ1

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Answer:

Step-by-step explanation:

League A                   League B

151.12                            163.25

148                                157

26.83                             24.93

29                                  136

136                                145

167                                178

207                               256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(82.5-1.5\times 76.56,159.06+1.5\times 76.56)

(-32.34,273.9)

So, There is no outlier

Maximum = 207

2)

League B in ascending order :

24.93,136,145,157,163.25,178,256

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(140.5-1.5\times 30.125,170.625+1.5\times 30.125)

(95.3125,215.8125)

24.93 and 256 are outliers  

Maximum = 256

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weeeeeb [17]

Step-by-step explanation:

1×1×1 |/£_《פ_《×~|_`×+₽€`_`+《_×~_《~×《_£~《,~|/€_|~£|_~|×~_《€52544151^1%1%1%!%^6!54151^6$16$17%61"`×<£《>_/£>_|×>|×`|_《~×《_《~×`《_¤《+_¤+》_{+|_`+|_{×~_+_+~£¤~\[£ зцсругтсурсшу

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0.4 / 17= what i need help this is ilx ​
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Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
We expect that students who do well on the midterm exam in a course will usually also do well on the final exam. Gary Smith of P
dybincka [34]

Using the model equation, the predicted mean score on the final given a score of <em>10 points above the class mean</em> in the mid term exam is 50.7

<u>The Least - Square Regression equation which models the relationship between midterm and final exam score is</u> :

  • γ=46.6 + 0.41x

x = 10 points ; <u>substitute the value of x = 10 into the regression equation</u> ;

γ=46.6 + 0.41(10)

γ=46.6 + 4.1

γ = 50.7

The <em>number of points above the mean</em> he'll score in the final exam is predicted to be 50.7

Learn more :brainly.com/question/18405415

7 0
3 years ago
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