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For a normal distribution, the area to the right hand side of the point x1x_1x1​ is 0.6 and to left hand side of the point x2x_2 x2​​  
Question

For a normal distribution, the area to the right hand side of the point x1x_1 is 0.6 and to left hand side of the point x2x_2 ​  is 0.7, then we have

A.

x1>x2x_1>x_2​​

B.

​​​​​​​​​​​​​​​x1<x2x_1<x_2​​

C.

​​​​​​​​​x1=x2 x_1=x_2​​

D.

None of these

Correct option is B

Given:
Area to the right of x1=610=0.6x_1=\frac{6}{10}=0.6​​​
Area to the left of x2=710=0.7x_2=\frac{7}{10}=0.7
Formula Used:
Area to the left of x1=1Area to the right of x1 x_1=1-\text{Area to the right of }x_1
Solution:
​Area to the left of x1=10.6=0.4. x_1=1-0.6=0.4.​​​
Area to the left of x2=0.7.x_2=0.7.​​​
Since 0.4<0.7, we have
x1<x2.x_1<x_2.​​​
The correct answer is (b) x1<x2. x_1<x_2.

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