Question

We made the shift from working with discrete random variables back in Unit 2 to now working with continuous random variables. Highly important for the rest of the course is our ability to use the Normal Distribution, in which we either find a probability from a range of values of the normally distributed variable X or we find a range of values from a given probability.

We will use shorthand notation (review LM 09 for how to use this notation) and probability notation for random variables (review LM 09 and LM 07 as needed) when working with normally distributed random variables. Suppose the vitamin C content of a particular variety of orange is distributed normally with mean 720 IU and standard deviation 46 IU. If we designate

X = the vitamin C content of a randomly selected orange,

then our shorthand notation is

X~N(720 IU, 46 IU).

Use this distribution of vitamin C content to answer the following questions:

3pt 1) What is the probability that a randomly selected orange will have less than 660 IU? Using X as the random variable, state your answer as a probability statement using the probability notation developed in the learning module.

3pt   2) What is the 80th percentile of the of the distribution of vitamin C content of the oranges?

1pt   3) What proportion of oranges exceed the vitamin C content you found in part (2) above?

3pt   4) What range of vitamin C content values represent the middle 80% of the distribution? State your answer as a probability statement using the probability notation developed in the learning module.

Extra Credit:

3pt EC   Suppose Y~N( 280 mg, 20 mg). Find Y₁ such that P( Y > Y₁) = 0.0250. State your answer in the form of a complete sentence without using any probability notation.

Answer 1

X~N(720 IU, 46 IU)., so mean=720 and variance(X)=46 and standard deviation(sd)=sqrt(variance)=sqrt(46)

we use standard normal variate Z=(X-mean)/sd=(X-720)/sqrt(46))

3pt 1) What is the probability that a randomly selected orange will have less than 660 IU? Using X as the random variable, state your answer as a probability statement using the probability notation developed in the learning module.

for X=660, Z=(660-720)/sqrt(46))=-8.85

required probability=P(X<660)=P(Z<-8.85)=0 ( using ms-excel=normsdist(-8.85))

3pt   2) What is the 80th percentile of the of the distribution of vitamin C content of the oranges?

first we find z such that  P(Z<z)=0.8 and z=0.8416

and required 80th percentile=x=mean+z*sd=720+0.8416*sqrt(46)=725.71

1pt   3) What proportion of oranges exceed the vitamin C content you found in part (2) above?

required proportion exceed=1-0.8416=0.1584

3pt   4) What range of vitamin C content values represent the middle 80% of the distribution? State your answer as a probability statement using the probability notation developed in the learning module.

we want to find x1 and x2 such that P(x1<X<x2)=0.8

for this we find z1 and z2 such that P(z1<Z<z2)=0.8

or P(Z<z2)-P(Z<z1)=0.8

or,1-P(Z<z1)-P(Z<z1)=0.8 ( since Z is symmetric)

or, P(Z<z1)=(1-0.8)/2=0.1 and z1=-1.2816 and z2=1.2816 ( since Z is symmetric about mean 0)

now x1=mean+z1*sd=720-1.2816*sqrt(46)=711.31

x2=720+1.2816*sqrt(46)=728.69

required range=(711.31,728.69)

Extra Credit:

3pt EC   Suppose Y~N( 280 mg, 20 mg). Find Y₁ such that P( Y > Y₁) = 0.0250. State your answer in the form of a complete sentence without using any probability notation.

here first we find such that P(Z>z1)=0.025 this imply P(Z<z1)=1-P(Z>z1)=1-0.025=0.975

and z1=1.96

and Y1=mean+z1*sd=280+1.96*sqrt(20))=288.77