Question

The weight of a 5th grader is normally distributed with a mean of 82 pounds and a variance of 76 pounds^2. Let weight in pounds, be represented by random variable X.

P(77<X< x2)=0.1070 Find x2.

Answer 1

TOPIC:Normal distribution.

- Define, x = The weight of a 5th graden student. (in bounds). Given that, TxvN (M = 82, 82 = 76) such that - We need to find 'X2' ; p (77.(x (x2) = 0.1070. Distribution curve :- dx (2) 77 IC 82 this H (Area of this region = 0.1070)

Now, we have, 76 p(77 < x < x2) 20:1070 ** (***82 < to be a VT759) 20:07 [ -=(80) (-0.5795 (2 < 2,1782) - 0-1070. I ~ N(O,D. [ 90) = cdy of 27 + (7-82 ) - 0 (-0.5733) = 0.1040, pp lacz<b) 7 123–82) – 21-4 (0.5435) 20-1040 . From 2- table7 >> (*782) - {1-0- 18% = 0.1070. 20:363) = (2-82) 0.3902 ( 76 L 2 (0.5735 20.168. 0.3902 Again, from z -table, 9(-0.279) 20. 3902 = - (-0.279). 76 / &, - 82 You can also use 'R' to find, 'k' when (1 = 0.39 02, = -0.279. √76 R-CODE - qnor - 0.279. norm (0.3902 =) X₂ 21 79. 5677 ) *. - 79.5677) (an). (in our decimals)