From the information give,
A) the probability that the amount paid by him is greater than 105 is ≈ 0.9332
B) the probability that the amount paid by him is not between 120 to 180 ≈ 0.3173
C) the probability that the amount paid by him is between 140 and 160 ≈ 0.2611
D) the probability that the amount paid by him is less than 180 ≈ 0.8413
E) the probability that the amount paid by him is between 140 and 160 given that greater than 120 ≈ 0.3104
F) the probability that the amount paid by him is between 120 and 160orbetween 140 to 180 ≈ 0.6827. See all computations below.
<h3>What is the calculation of for the solutions indicated above?</h3>
Let X be the amount that the student paid;
Note that X ∼ N(μ,σ² ).
Since we have μ = 150, and σ = 30; The1refore
A) P(X>105)=1−P(X≤105)
= 1 - P (Z ≤ (105 -150)/30)
= - P (Z ≤ -1.5)
P(X>105) ≈ 0.9332
B) P(X<120 or X>180)
=P(X<120)+1−P(X≤180)
= P ((Z < (120 - 150)/30) + 1 - P ((Z < (180 - 150)/30)
=P(Z<−1)+1−P(Z≤1)
≈0.158655+0.158655
P(X<120 or X>180) ≈ 0.3173
C) P(140<X<160)
= P(X<160)−P(X≤140)
= P ((Z < (160 - 150)/30) + 1 - P ((Z < (140 - 150)/30)
≈P(Z<0.33333)−P(Z≤−0.33333)
≈0.63056−0.36944
P(140<X<160) ≈ 0.2611
D) P(X<180) = P(Z< (180−150)/30
=P(Z<1)
P(X<180) ≈0.8413
E) P(140<X<160∣X>120)
= (P((140<X<160)∩(X>120))/P(X>120)
= (P((140<X<160))/P(X>120)
= (P Z < (160 -150)/30) - P (Z ≤ (140-150)/30))/ 1- P (Z ≤ (120 - 150)/30)
≈ (P(Z<0.33333)−P(Z≤−0.33333))/ 1 - P(Z ≤ - 1)
≈ (0.63056−0.36944)/0.84134
P(140<X<160∣X>120) ≈ 0.3104
F) P(120<X<160 or 140<X<180)
= P(120<X<180)
= P(X<180)−P(X≤120)
= P ((Z < (180 -150)/30) - P ((Z ≤ (120 -150)/30)
= P(Z<1)−P(Z≤−1)
≈ 0.841345−0.158655
P(120<X<160 or 140<X<180) ≈ 0.6827
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