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Darina [25.2K]
4 years ago
10

On the last three physics exams a student scored 89, 87, and 92. What score must the student earn on the next exam to have an av

erage above 90?
Lots represent the score the student must eam What is this score?
S>
(Simplify your answer. Type an intogor or a decimal)
Mathematics
1 answer:
OlgaM077 [116]4 years ago
4 0

Answer:

For the average to be above 90, the marks in the fourth paper should be MORE THAN 92.

Step-by-step explanation:

The score of students in last 3 exams =  89 , 87 and 92

Let us assume the marks scored by the student in last exam = k

The average of marks should be above 90.

Now, \textrm{The average of total marks}  = \frac{\textrm{Sum of all marks obtained in 4 papers}}{\textrm{ 4 papers }}

Sum of all 4 paper's marks =  89 + 87 + 92 + k = 268 + k

Also, 90 <  Average

\implies 90 < \frac{268 + k}{4}\\ or, 360 < 268 + k\\\implies  360 - 268 < k\\or,  k >  92

Hence,for the average to be above 90, the marks in the fourth paper should be MORE THAN 92.

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To find the solution we will first find the difference between each number and mean, and then substitute the value in the formula of standard deviation.

The standard deviation is 8.4477 and the variance is 71.40.

Given to us

  • Test scores of Adimas = (76, 87, 65, 88, 67, 84, 77, 82, 91, 85, 90)
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<h3>Standard Deviation</h3>

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\sum(x-\mu)^2 = x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2+x_8^2+x_9^2+x_{10}^2+x_{11}^2

                 =  25 + 36 + 256 + 49 + 196 + 9 + 16 + 1 + 100 +16 + 81

                 = 785

\rm{ Standard\ Deviation = \sqrt{\dfrac{\sum{(X-\mu)^2}} {n}

                             \sigma= \sqrt{\dfrac{785}{11}}\\\\&#10;\sigma = 8.4477

<h3>Variance</h3>

Variance = \sigma ^2

              = 71.40

Hence, the standard deviation is 8.4477 and the variance is 71.40.

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