Question #99993
mr. jansen gives a unit test to his class. he determines that the mean mark on the test is 62% and the standard deviation is 8 marks. when he hands the tests back a student realized that one question is wrong and is actually impossible to solve. so the teacher adds 5 percent to all the students marks, what is the mean and standard deviation of the new set of marks?
1
Expert's answer
2019-12-06T09:49:19-0500
μ=1ni=1nxi\mu={1\over n}\displaystyle\sum_{i=1}^nx_iσ2=1ni=1n(xiμ)2\sigma^2={1\over n}\displaystyle\sum_{i=1}^n(x_i-\mu)^2

σ=σ2\sigma=\sqrt{\sigma^2}

μ1=1ni=1nxi1=62\mu_1={1\over n}\displaystyle\sum_{i=1}^nx_{i1}=62

σ12=1ni=1n(xi1μ1)2=82\sigma_1^2={1\over n}\displaystyle\sum_{i=1}^n(x_{i1}-\mu_1)^2=8^2

Given that

xi2=xi1+5, i=1,2,...,nx_{i2}=x_{i1}+5,\ i=1,2,...,n

Then


μ2=1ni=1nxi2=\mu_2={1\over n}\displaystyle\sum_{i=1}^nx_{i2}==1ni=1n(xi1+5)=={1\over n}\displaystyle\sum_{i=1}^n(x_{i1}+5)==1ni=1nxi1+5=μ1+5=62+5=67={1\over n}\displaystyle\sum_{i=1}^nx_{i1}+5=\mu_1+5=62+5=67

σ22=1ni=1n(xi2μ2)2=\sigma_2^2={1\over n}\displaystyle\sum_{i=1}^n(x_{i2}-\mu_2)^2==1ni=1n((xi1+5)(μ1+5))2=={1\over n}\displaystyle\sum_{i=1}^n((x_{i1}+5)-(\mu_1+5))^2==1ni=1n(xi1μ1)2=σ12=82={1\over n}\displaystyle\sum_{i=1}^n(x_{i1}-\mu_1)^2=\sigma_1^2=8^2



The mean will increase by 5 (it will be 6767).

The standard deviation will remain the same (it will be 88).




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