There are six orange marbles, 2 red marbles, 3 white marbles, and 4 green marbles in a bag. Once a marble is drawn, it is replac
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Answer:
(a) P(0 ≤ Z ≤ 2.87)=0.498
(b) P(0 ≤ Z ≤ 2)=0.477
(c) P(−2.20 ≤ Z ≤ 0)=0.486
(d) P(−2.20 ≤ Z ≤ 2.20)=0.972
(e) P(Z ≤ 1.01)=0.844
(f) P(−1.95 ≤ Z)=0.974
(g) P(−1.20 ≤ Z ≤ 2.00)=0.862
(h) P(1.01 ≤ Z ≤ 2.50)=0.150
(i) P(1.20 ≤ Z)=0.115
(j) P(|Z| ≤ 2.50)=0.988
Step-by-step explanation:
(a) P(0 ≤ Z ≤ 2.87)
In this case, this is equal to the difference between P(z<2.87) and P(z<0). The last term is substracting because is the area under the curve that is included in P(z<2.87) but does not correspond because the other condition is that z>0.
(b) P(0 ≤ Z ≤ 2)
This is the same case as point a.
(c) P(−2.20 ≤ Z ≤ 0)
This is the same case as point a.
(d) P(−2.20 ≤ Z ≤ 2.20)
This is the same case as point a.
(e) P(Z ≤ 1.01)
This can be calculated simply as the area under the curve for z from -infinity to z=1.01.
(f) P(−1.95 ≤ Z)
This is best expressed as P(z≥-1.95), and is calculated as the area under the curve that goes from z=-1.95 to infininity.
It also can be calculated, thanks to the symmetry in z=0 of the standard normal distribution, as P(z≥-1.95)=P(z≤1.95).
(g) P(−1.20 ≤ Z ≤ 2.00)
This is the same case as point a.
(h) P(1.01 ≤ Z ≤ 2.50)
This is the same case as point a.
(i) P(1.20 ≤ Z)
This is the same case as point f.
(j) P(|Z| ≤ 2.50)
In this case, the z is expressed in absolute value. If z is positive, it has to be under 2.5. If z is negative, it means it has to be over -2.5. So this probability is translated to P|Z| < 2.50)=P(-2.5<z<2.5) and then solved from there like in point a.
