Question

For normal distribution with mean as 500 and SD as 120,What is the value of k so that the interval (500, k) covers 40.32 per sent area of the normal curve?

Answer 1

K = 656  or 344  if For normal distribution  mean 500 and SD 120 and area covered between (500 , K) = 40.32%

Step-by-step explanation:

Mean = 500

SD = 120

500 , k

Z =  (Value - Mean)/SD

=>  Z = (500 - 500)/120

=> Z = 0

=> 50% area is before 500

to Cover 40.32 % area

either K should cover   50 + 40.32 = 90.32 %

or 50 - 40.32 = 9.68%

Z for 90.32 % =  1.3

Z for 9.68% % =  -1.3

1.3  = ( K - 500)/120

=> K = 156 + 500

=> K = 656

or

-1.3  = ( K - 500)/120

=> K = -156 + 500

=> K = 344

(344 , 500 )   or ( 500 , 656) will cover 40.32% area

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