Time spent using email per session is normally distributed with an average of 9

Question

Time spent using email per session is normally distributed with an average of 9 minutes and standard deviation of 2 minutes.                                    

a) In a sample of 25 sessions, what is the probability the sample mean is between 8.8 and 9.2 minutes? (round to 3 decimal places)

b) In a sample of 25 sessions, what is the probability the sample mean is between 8.5 and 9 minutes? (round to 3 decimals)

c) In a sample of 200 sessions, what is the probability the sample mean will be between 8.8 and 9.2 minutes? (round to 3 decimals)

d) What effect does sample size have on the standard error of the mean?

Explanation / Answer

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    8.8      
x2 = upper bound =    9.2      
u = mean =    9      
n = sample size =    25      
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.5      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.308537539      
P(z < z2) =    0.691462461      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.382924923   [ANSWER]

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

b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    8.5      
x2 = upper bound =    9      
u = mean =    9      
n = sample size =    25      
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.25      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.105649774      
P(z < z2) =    0.5      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.394350226   [ANSWER]

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

c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    8.8      
x2 = upper bound =    9.2      
u = mean =    9      
n = sample size =    200      
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.414213562      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.414213562      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.078649604      
P(z < z2) =    0.921350396      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.842700793   [ANSWER]

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

d)

As SE = s/sqrt(n), then we see that as sample size increases, standard error decreases.  
  
  

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