Given a normal population whose mean is 50 and whose standar

Given a normal population whose mean is 50 and whose standard deviation is 10.

a. Find the probability that a random sample of 4 has a mean between 49 and 52. (to 4 decimals)

b. Find the probability that a random sample of 16 has a mean between 49 and 52. (to 4 decimals)

c. Find the probability that a random sample of 25 has a mean between 49 and 52. (to 4 decimals)

Solution

a)
Mean ( u ) =50
Standard Deviation ( sd )=10
Number ( n ) = 4
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 49) = (49-50)/10/ Sqrt ( 4 )
= -1/5
= -0.2
= P ( Z <-0.2) From Standard Normal Table
= 0.42074
P(X < 52) = (52-50)/10/ Sqrt ( 4 )
= 2/5 = 0.4
= P ( Z <0.4) From Standard Normal Table
= 0.65542
P(49 < X < 52) = 0.65542-0.42074 = 0.2347                  

b)
WHEN n = 16
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 49) = (49-50)/10/ Sqrt ( 16 )
= -1/2.5
= -0.4
= P ( Z <-0.4) From Standard Normal Table
= 0.34458
P(X < 52) = (52-50)/10/ Sqrt ( 16 )
= 2/2.5 = 0.8
= P ( Z <0.8) From Standard Normal Table
= 0.78814
P(49 < X < 52) = 0.78814-0.34458 = 0.4436                  

c)
When n= 25
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 14) = (14-18)/3.8419
= -4/3.8419 = -1.0412
= P ( Z <-1.0412) From Standard Normal Table
= 0.1489
P(X < 22) = (22-18)/3.8419
= 4/3.8419 = 1.0412
= P ( Z <1.0412) From Standard Normal Table
= 0.8511
P(14 < X < 22) = 0.8511-0.1489 = 0.7022