madisonhemmo7
Answered

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Another distribution problem. Can someone solve?

Another Distribution Problem Can Someone Solve class=

Sagot :

caylus

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 .

madethisfortwitch1

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

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