The graph shows the distance y, in centimeters, a pendulum moves to the right (positive displacement) and to the left (negative
displacement), for a given number of seconds x.
How many seconds are required for the pendulum to swing from its position furthest to the right to its position furthest to the left?
How many seconds are required for the pendulum to swing from its position furthest to the right to its position furthest to the left?
1 answer:
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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.
