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Sagot :
The probabilities in the question are
- 0.0062
- 0.7499
- 0.000429
How to solve for the probabilities
a. For x < 70
we have
z< 70 - 100/12
= z < -30/12
= -2.5
Such that p (x<70) = 0.0062
Hence the probability that is is less than $70 = 0.0062
b. between $90 and $120,
90 - 100/12. 120 - 100/12
= -0.8333 <z< 1.67
p(90<x<120) = 0.95224 - 0.20234
= 0.7499
0.7499 is the probability of between $90 and $120.
c. more than $140
140-100/12
= P(Z>3.3333)
= 0.000429
Read more on probability here:
https://brainly.com/question/24756209
#SPJ1
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