a Suppose a random sample of n 100 is sampled from a popula

a) Suppose a random sample of n = 100 is sampled from a population with mean mu = 65 and sigma = 40, Find P(bar x

Solution

Normal Distribution
Mean ( u ) =65
Standard Deviation ( sd )=40
Number ( n ) = 100
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X < 62) =   (62-65)/40/ Sqrt ( 100 )
= -3/4= -0.75
= P ( Z <-0.75) From Standard NOrmal Table
= 0.2266                  
b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 62) = (62-65)/40/ Sqrt ( 100 )
= -3/4
= -0.75
= P ( Z <-0.75) From Standard Normal Table
= 0.22663
P(X < 68) = (68-65)/40/ Sqrt ( 100 )
= 3/4 = 0.75
= P ( Z <0.75) From Standard Normal Table
= 0.77337
P(62 < X < 68) = 0.77337-0.22663 = 0.5467                  
II)
Normal Distribution
Mean ( np ) =80
Standard Deviation ( ?npq )= ?4000*0.02*0.98 = 8.8544
Normal Distribution = Z= X- u / sd                   
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 89.5) = (89.5-80)/8.8544
= 9.5/8.8544 = 1.0729
= P ( Z <1.0729) From Standard Normal Table
= 0.85834
P(X < 90.5) = (90.5-80)/8.8544
= 10.5/8.8544 = 1.1859
= P ( Z <1.1859) From Standard Normal Table
= 0.88216
P(89.5 < X < 90.5) = 0.88216-0.85834 = 0.0238                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 75) = (75-80)/8.8544
= -5/8.8544 = -0.5647
= P ( Z <-0.5647) From Standard Normal Table
= 0.28614
P(X < 86) = (86-80)/8.8544
= 6/8.8544 = 0.6776
= P ( Z <0.6776) From Standard Normal Table
= 0.751
P(75 < X < 86) = 0.751-0.28614 = 0.4649