assume data is normally distributed give answers to three de

(assume data is normally distributed, give answers to three decimal places)

Solution

Mean ( u ) =90
Standard Deviation ( sd )=23.4
Number ( n ) = 45
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X < 86) = (86-90)/23.4/ Sqrt ( 45 )
= -4/3.4883= -1.1467
= P ( Z <-1.1467) From Standard NOrmal Table
= 0.126                
b)
P(X > 110) = (110-90)/23.4/ Sqrt ( 45 )
= 20/3.488= 5.7335
= P ( Z >5.7335) From Standard Normal Table
= 0                  
c)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 99.5) = (99.5-90)/23.4/ Sqrt ( 45 )
= 9.5/3.4883
= 2.7234
= P ( Z <2.7234) From Standard Normal Table
= 0.99677
P(X < 100.5) = (100.5-90)/23.4/ Sqrt ( 45 )
= 10.5/3.4883 = 3.0101
= P ( Z <3.0101) From Standard Normal Table
= 0.99869
P(99.5 < X < 100.5) = 0.99869-0.99677 = 0.002