Probability

Question

In four tests taken by 450 students, marks are found to be normally distributed with mean and variance as given below

Test Id	Mean	Variance
1	74	121
2	75	100
3	78	196
4	82	169

 

A has secured 80 in the first test, 81 in the second, 86 in the third and 89 in the fourth. In which test did A actually perform best relative to other students.

(A)Fourth Test

(B)Third Test

(C)Second Test

(D)First Test

 

Comments

I derived standard normal values of A's score in first,second,third and fourth tests respectively and they are

$0.5454,0.60,0.571,0.5384$

$Z=\frac{X-\mu}{\sigma}$

So, according to me answer must be second test.

Am I correct?

Answer 1

 In which test did A actually perform best relative to other students?

To find out we have to calculate the performance of student in comparison to other students appearing in the same test

In Test 1,

$\mu _1$ = 74 , $\sigma_1 = 11$

$Z_1 = \frac{X_1 - \mu_1}{\sigma_1} =  \frac{80 - 74}{11} = 0.545 $

Area to the left of Z is 0.705402 so the student performed better than $70.54 \%$ students in test1.

$\mu _2$ = 75 , $\sigma_2 = 10$

$Z_2 = \frac{X_2 - \mu_2}{\sigma_2} =  \frac{81 - 75}{10} = 0.6$

Area to the left of Z is 0.725747 so the student performed better than $72.57 \%$ students in test2.

$\mu _3$ = 78 , $\sigma_3 = 14$

$Z_3 = \frac{X_3 - \mu_3}{\sigma_3} =  \frac{86 - 78}{14} = 0.571$

Area to the left of Z is 0.715661 so the student performed better than $71.56 \%$ students in test3.

$\mu _4$ = 82 , $\sigma_4 = 13$

$Z_4 = \frac{X_4 - \mu_4}{\sigma_4} =  \frac{89 - 82}{13} = 0.538$

Area to the left of Z is 0.701944 so the student performed better than $70.19 \%$ students in test4.

So from this we can conclude that student performed better in 2nd Test relative to other students.

 

Comments

Area to the left of Z is 0.705402  what do u mean by area to left

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Deepanshu For this particular question it is not necessary to calculate area under curve because student performed more than mean in every test. 

what do u mean by area to left

Area to the left of left of Z value in Normal curve. 

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Deepanshu i just mentioned the area under curve for reader to understand why anyone is calculating Z score and how Z score is helping in calculating percentile of student. 

Area to the left of Z tell us how many % of people have scored less than the student in a given test

Area to the right of Z tell us how many % of people have scored more than the student in a given test

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Mk Utkarsh not that much good in probability ......

i left this topic due to that big formula ( i  remember ) plzz share good link of this topic ........if u have any ......

 

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that formula will never be used don't worry, watch khan academy videos and solve questions from hk dass book

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@Mk  Utkarsh Please tell how to find area?

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Shadan Karim If area is required to know the answer then it will be given in the question otherwise if it is not required like in this question it will not be given because for this question with only Z score we can determine the answer.

Answer 2

Graphical interpretation:

If you convert the data points to standard normal form then the graph will always be symmetrical as shown below. Now each normal form that we have calculated in the above answer can be represented as:

So when do we get a better rank? When more people have scored less than you. That means in whichever test’s data-point you will have more area on the left-hand side means more people scored less than you and you performed better, and we can see that in TEST-2 we have more area on the left-hand side.