Question

1. Suppose that the number of 12 oz sodas consumed in a week by an 18 -22 year old is normally distributed with distribution N(12, 4.5).   How small can the mean of a sample of size forty be and still be in the top 10%? Round to two decimal places.

2. Suppose that the number of 12 oz sodas consumed in a week by an 18 -22 year old is normally distributed with distribution N(12, 4.5). How small can the mean of a sample of size forty be and still be in the middle 90%? Round to two decimal places.

3.

Now we will switch populations. We are looking at the number of sodas that a 40 - 50 year old consumes in a week. We know this population is normal. We are pretty sure that $LaTeX: \sigma$ σ = 4.5 (we will use that) but we do not know $LaTeX: \mu$ μ.  

If we randomly select a sample of size n=100 from this population and take the mean, what is the probability they average more than $LaTeX: \mu+\:.45$ μ+.45 sodas in a week. Round to two places.

Answer 1

1)

µ =    12                              
σ =    4.5                              
n=   40                              
proportion=   0.9000                              
                                  
Z value at    0.9   =   1.282   (excel formula =NORMSINV(   0.90   ) )          
z=(x-µ)/(σ/√n)                                  
so, X=z * σ/√n +µ=   1.282   *   4.5   / √    40   +   12   = 12.91

2)

µ =    12                              
σ =    4.5                              
n=   40                              
proportion=   0.9                              
proportion left    0.1   is equally distributed both left and right side of normal curve                           
z value at   0.05   = ±   -1.64   (excel formula =NORMSINV(   0.10   / 2 ) )          
Z value at    0.95   =   1.64                      
z = ( x - µ ) / (σ/√n)                                  
so, X = z σ / √n + µ =                                  
X1 =   -1.64   *   4.5   / √   40   +   12   =   10.83
X2 =   1.64   *   4.5   / √   40   +   12   =   13.17

3)

σ =    4.5                                      
n=   40                                      
                                          

                                          
Z =   (X - µ )/(σ/√n) = ( 0.45) / (    4.5   / √   40   ) =   0.632  
                                          
P(X-µ ≥ 0.45   ) = P(Z ≥   0.63   ) =   P ( Z <   -0.632   ) =    0.2635 or 26.35% (answer)