Question

1. Random variable X is drawn from a normal distribution with mean 5.44 and std dev 2.54.
   1. Calculate the probability of X being less than 3.29.
   2. What is the probability of X exceeding 4.61?
   3. What is the probability of X lying between 5.79 and 7.8?
   4. Verify your answers to parts 1 2 and 3 above using numerical sampling.

Answer 1

Mean = 5.44

Standard deviation = 2.54

a) Probability of X being less than 3.29:

Z = ( X – μ ) / σ

= ( 3.29 – 5.44) / 2.54

Z = - 0.846

              Probability of X being less than 3.29 = [ Area to the right of Z = 0 ] – [ Area between Z=0 to Z = 0.846 ]

                                                                                        = 0.5 – 0.2995

                                                                                        = 0.2005

                                                                                       

b) Probability of X exceeding 4.61:

Z = ( X – μ ) / σ

Z = ( 4.61 – 5.44) / 2.54

             Z = -0.326

[ Probability of X exceeding 4.61 ] = [ Area to the right of Z = 0 ] – [ Area between Z=0 to Z = 0.326 ]

                                                                   = 0.5 – 0.1255

                                                                   = 0.3745

c) Probability of X lying between 5.79 and 7.8:

For x= 5.79

Z = ( X – μ ) / σ

Z = ( 5.79 – 5.44) / 2.54

Z = 0.137

For x= 7.8

Z = ( X – μ ) / σ

Z = ( 7.8 – 5.44) / 2.54

Z = 0.929

Probability of X lying between 5.79 and 7.8 = [ Area from Z =0 to Z=0.929 ] - [Area from Z=0 to Z=0.137]

= 0.3212 - 0.0517

= 0.2695