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Suppose that the mean and s.d of the tuition fee paid by bs mathematics students in umt is 150and 30 in uds, respectively. it is assumed that data is normally distributed. if a student is selected at random, find the probability that the amount paid by him is
i) greater than 105
ii) not between 120 to 180
iii) between 140 and 160
iv) less than 180
v) between 140 and 160 given that greater than 120
vi) between 120 and 160orbetween 140 to 180

Respuesta :

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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.

What is the calculation of for the solutions indicated above?

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

Learn more about probability at:

https://brainly.com/question/24756209
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