assume sigma 2 If n 9 bottles are randomly selected what i

assume sigma = 2. If n = 9 bottles are randomly selected, what is P(|Y - mu

Solution

a)

Note that

|Ybar - u| < 0.3

|Ybar - u|*sqrt(n)/sigma < 0.3*sqrt(n)/sigma

|Ybar - u|*sqrt(n)/sigma < 0.3*sqrt(9)/2

|z| < 0.45

-0.45 < z < 0.45.

z1 = lower z score =    -0.45      
z2 = upper z score =     0.45      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.32635522      
P(z < z2) =    0.67364478      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.347289559   [ANSWER]

************************

B)

For n = 25:

|Ybar - u|*sqrt(n)/sigma < 0.3*sqrt(25)/2

|z| < 0.75

-0.75 < z < 0.75.

z1 = lower z score =    -0.75      
z2 = upper z score =     0.75      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.226627352      
P(z < z2) =    0.773372648      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.546745295      

....

For n = 36:
  
|Ybar - u|*sqrt(n)/sigma < 0.3*sqrt(36)/2

|z| < 0.9

-0.9 < z < 0.9.

z1 = lower z score =    -0.9      
z2 = upper z score =     0.9      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.184060125      
P(z < z2) =    0.815939875      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.631879749      

....

For n = 49:

|Ybar - u|*sqrt(n)/sigma < 0.3*sqrt(49)/2

|z| < 1.05

-1.05 < z < 1.05.

z1 = lower z score =    -1.05      
z2 = upper z score =     1.05      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.146859056      
P(z < z2) =    0.853140944      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.706281887      

....

For n = 64:

|Ybar - u|*sqrt(n)/sigma < 0.3*sqrt(64)/2

|z| < 1.2

-1.2 < z < 1.2

z1 = lower z score =    -1.2      
z2 = upper z score =     1.2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.11506967      
P(z < z2) =    0.88493033      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.76986066      

**************************************************

c)

Thus, as n increases, the probability increases.

**************************************

d)

The probabilities here are smaller.