Another distribution problem. Can someone solve?

Answer:
Hello,
Step-by-step explanation:
[tex]P(3000<x<4000)\\\\=P(\dfrac{3000-3262}{1100} \leq z\leq \dfrac{4000-3262}{1100})\\\\=P(z\leq 0,6709)-(1-P(z\leq 3524) \leq z))\\\\=0.7515-0.5941\\\\=0.1574\\[/tex]
My table have 4 digits but i have made a linear interpolation .
Answer:
-.238, .67, .7486, .4052, .3434
Step-by-step 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