Question #207546

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


1
Expert's answer
2021-06-16T17:05:36-0400

Let X=X= the amount paid by student; XN(μ,σ2).X\sim N(\mu, \sigma^2).

Given μ=150,σ=30\mu=150, \sigma=30

i)


P(X>105)=1P(X105)P(X>105)=1-P(X\leq105)

=1P(Z10515030)=1P(Z1.5)=1-P(Z\leq\dfrac{105-150}{30})=1-P(Z\leq-1.5)

0.9332\approx0.9332

ii)


P(X<120 or X>180)P(X<120\ or\ X>180)

=P(X<120)+1P(X180)=P(X<120)+1-P(X\leq180)

=P(Z<12015030)+1P(Z18015030)=P(Z<\dfrac{120-150}{30})+1-P(Z\leq\dfrac{180-150}{30})

=P(Z<1)+1P(Z1)=P(Z<-1)+1-P(Z\leq1)

0.158655+0.1586550.3173\approx0.158655+0.158655\approx0.3173

iii)


P(140<X<160)P(140<X<160)

=P(X<160)P(X140)=P(X<160)-P(X\leq140)

=P(Z<16015030)P(Z14015030)=P(Z<\dfrac{160-150}{30})-P(Z\leq\dfrac{140-150}{30})

P(Z<0.33333)P(Z0.33333)\approx P(Z<0.33333)-P(Z\leq-0.33333)

0.630560.369440.2611\approx0.63056-0.36944\approx0.2611

iv)


P(X<180)=P(Z<18015030)P(X<180)=P(Z<\dfrac{180-150}{30})

=P(Z<1)0.8413=P(Z<1)\approx0.8413

v)


P(140<X<160X>120)P(140<X<160|X>120)

=P((140<X<160)(X>120))P(X>120)=\dfrac{P((140<X<160)\cap (X>120))}{P(X>120)}

=P(140<X<160)P(X>120)=\dfrac{P(140<X<160)}{P(X>120)}

=P(Z<16015030)P(Z14015030)1P(Z12015030)=\dfrac{P(Z<\dfrac{160-150}{30})-P(Z\leq\dfrac{140-150}{30})}{1-P(Z\leq\dfrac{120-150}{30})}

P(Z<0.33333)P(Z0.33333)1P(Z1)\approx\dfrac{P(Z<0.33333)-P(Z\leq-0.33333)}{1-P(Z\leq-1)}

0.630560.369440.841340.3104\approx\dfrac{0.63056-0.36944}{0.84134}\approx0.3104

vi)


P(120<X<160 or 140<X<180)P(120<X<160\ or\ 140<X<180)

=P(120<X<180)=P(120<X<180)

=P(X<180)P(X120)=P(X<180)-P(X\leq120)

=P(Z<18015030)P(Z12015030)=P(Z<\dfrac{180-150}{30})-P(Z\leq\dfrac{120-150}{30})

=P(Z<1)P(Z1)=P(Z<1)-P(Z\leq-1)

0.8413450.1586550.6827\approx0.841345-0.158655\approx0.6827


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