A consumer advocacy group tested the onair lifetimes a rando

A consumer advocacy group tested the \"on-air\" lifetimes a random sample of 162 cell phone batteries. The mean lifetime was 2.9 hours with a standard deviation of 0.4 hours. The lifetimes are approximately bell-shaped. Estimate the number of batteries with lifetimes between 2.1 hours and 3.7 hours.

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    2.1      
x2 = upper bound =    3.7      
u = mean =    2.9      
          
s = standard deviation =    0.4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.954499736      

Thus, there are 0.954499736*162 = 154.6289572 or about 155 such batteries. [ANSWER]