Question

Another distribution problem. Can someone solve?

Answer 1

-.238, .67, .7486, .4052, .3434

Explanation:

standardize them both

(3000-3262)/1100= -.238 which rounds to -.24 which has a probability of .4052

(4000-3262)/1100= .67 which has a probability of .7486

subtract them

.7486-.4052=.3434

Answer 2

Hello,

Explanation:

`P(3000<x<4000)\\\\=P((3000-3262)/(1100) \leq z\leq (4000-3262)/(1100))\\\\=P(z\leq 0,6709)-(1-P(z\leq 3524) \leq z))\\\\=0.7515-0.5941\\\\=0.1574\\`

My table have 4 digits but i have made a linear interpolation .