7.3 APPLICATIONS OF THE NORMAL DISTRIBUTION

PROBABILITIES We want to calculate probabilities and values for general normal probability distributions We can relate these problems to calculations for the standard normal in the previous section

Z-SCORE For a general normal random variable X with mean μ and standard deviation σ, the variable has a standard normal probability distribution We can use this relationship to perform calculations for X

Z-SCORE Values of X  Values of Z If x is a value for X, then is a value for Z This is a very useful relationship

EXAMPLE ● For example, if  μ = 3  σ = 2 then a value of x = 4 for X corresponds to a value of z = 0.5 for Z

RELATIONSHIPS ●Because of this relationship Values of X  Values of Z Then P(X < x) = P(Z < z) ●To find P(X < x) for a general normal random variable, we could calculate P(Z < z) for the standard normal random variable

Z IS X ●With this relationship, the following method can be used to compute areas for a general normal random variable X  Shade the desired area to be computed for X  Convert all values of X to Z-scores using  Solve the problem for the standard normal Z (Section 7.2 Stuff)  The answer will be the same for the general normal X

EXAMPLE ●For a general random variable X with  μ = 3  σ = 2 calculate P(X < 6) ● This corresponds to so P(X < 6) = P(Z < 1.5) = Hint…Normalcdf(small, z, 0,1)

EXAMPLE ●For a general random variable X with  μ = –2  σ = 4 calculate P(X > –3 ) ● This corresponds to so P(X > –3 ) = P(Z > –0.25 ) = Hint…(right hand) Normalcdf(z, big number,0,1)

BETWEEN ●For a general random variable X with  μ = 6  σ = 4 calculate P(4 < X < 11) ● This corresponds to so P(4 < X < 11) = P( – 0.5 < Z < 1.25) = Hint…Normalcdf(smaller z,bigger z,0,1)

INVERSE The inverse of the relationship is the relationship With this, we can compute value problems for the general normal probability distribution

INVERSE ●The following method can be used to compute values for a general normal random variable X  Shade the desired area to be computed for X  Find the Z-scores for the same probability problem  Convert all the Z-scores to X using  This is the answer for the original problem

EXAMPLE ●For a general random variable X with  μ = 3  σ = 2 find the value x such that P(X < x) = 0.3 Find the Z that corresponds to.3 (InvNorm,.3,0,1) = ● Since P(Z < –0.5244) = 0.3, we calculate so P(X < 1.95) = P(Z < –0.5244) = 0.3

NOTICE THE GREATER THAN!! ●For a general random variable X with  μ = –2  σ = 4 find the value x such that P(X > x ) = 0.2 Since it is a “Righthand” we must use or.8 in our invNorm(.8,0,1) = so P(X > 1.37 ) = P(Z > ) = 0.2

SAME OLD Tough Stuff…work hard on this!