Question #161471

The weights of the students in a certain year level are normally distributed with a mean of 60kg and a standard deviation of 3.5kg. Find the probability that a student randomly selected from this group weighs. i) 55kg or less. ii) 68kg or more. iii) between 48kg and 53kg




1
Expert's answer
2021-02-24T07:58:38-0500

The weight of the students in a certain year level are normally distributed with mean 60kg60kg and standard deviation 3.5kg3.5kg .

Let XX be random variable which defines weight of the students.

Then , μ=60\mu =60 and σ=3.5\sigma =3.5

Let us take Z=XμσZ=\frac{X-\mu}{\sigma } . Then Z=X603.5Z=\frac{X-60}{3.5}

(i) Probability that a student randomly selected is less than or equal to 5555 kg is P(X55)P(X\leq 55) .

P(X55)=P(Z55603.5)\therefore P(X\leq 55)=P(Z \leq \frac{55-60}{3.5})

=P(Z1.43)=P(Z \leq -1.43)

=0.5P(0<Z<1.43)=0.5-P(0<Z<1.43)

=0.50.4236=0.5-0.4236 [ from normal distribution table]

=0.076=0.076 (approximately)

(ii) Probability that a student randomly selected is greater than or equal to 6868 kg is P(X68)P(X\geq 68) .

P(X68)=P(Z68603.5)\therefore P(X\geq 68)=P(Z \geq \frac{68-60}{3.5})

=P(Z2.29)=P(Z\geq 2.29)

=0.5P(0<Z<2.29)=0.5-P(0<Z<2.29)

=0.50.4889=0.5-0.4889

=0.011=0.011 (approximately)

(iii)Probability that a student randomly selected is greater than 4848 kg and less than 53 kg is P(48<X<53)P(48<X<53)

P(48<X<53)=P(48603.5<Z<53603.5)\therefore P(48<X<53)=P(\frac{48-60}{3.5}<Z<\frac{53-60}{3.5})

=P(3.43<Z<2)=P(-3.43<Z<-2)

=P(0<Z<3.43)P(0<Z<2)=P(0<Z<3.43)-P(0<Z<2)

=0.49960.4772=0.4996-0.4772

=0.022=0.022 (approximately)



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