ms-08 mba assignment july dec 2010 Question 3

3.From a set of 1000 observations known to be normally distributed, the mean is 534 cm and  SD is 13.5 cm.  How many observations are likely to exceed 561 cm? How many will be between 520.5 cm and 547.5 cm? Between what limits will the

middle 50% of the observations lie?

Solution:

Given: s = 13.5cm ,  m = 534cm , n = 1000

(i)                              P (X > 561)

Z = X - m      = (561 - 534) / 13.5 = 25/ 13.5 = 1.85

        s

P (X>561) = P ( Z> 1.85)

                  = 0.5- P (0 < Z < 1.85)

                  = 0.5 – 0.4744   (From the Table)

                  = 0.0256

Number of observations above 561 = 1000*0.0256 = 25.6

                                                      = 26cms (approx)

(ii)       P (520.5<X<547.5)

             Z = (520.5 – 534) / 13.5

                = -13.5/13.5 = -1

             Z= (547.5 – 534) / 13.5

               =  13.5/ 13.5 = 1

          P(-1< Z< 1) = P (-1< Z< 0) + P(0 < Z< 1)

                              = 2* P (0< Z< 1) = 2* 0.3413 = 0.6826

        Number of observations between 520.5 and 547.5 = 1000* 0.6826 = 682.6 = 683cms (approx)

(ii)                            Given 50% of middle value i.e Z = 0.5 on both sides of 0

i.e Z= 0.5/2 = 0.25 on one side of normal graph

The table value for 0.25 is 0.67

Confidence interval

                      m = Mean k0.67* s

                          = 534 k  0.67* 13.5

                         = 524.955, 543.045

The interval is from 524.955 to 543.045 where 50% of middle observations lie.