Average talk time between charges of a cell phone is adverti

Average talk time between charges of a cell phone is advertised as 4.3 hours. Assume that talk time is normally distributed with a standard deviation of 1.2 hour. a. Find the probability that talk time between charges for a randomly selected cell phone is below 3.1 hours. (Round \"z\" value to 2 decimal places and final answer to 4 decimal places.) b. Find the probability that talk time between charges for a randomly selected cell phone is either more than 5 hours or below 1.9 hours. (Round \"z\" value to 2 decimal places and final answer to 4 decimal places.) c. Fifty percent of the time, talk time between charges is below a particular value. What is this value? (Round \"z\" value to 2 decimal places and final answer to 3 decimal places.)

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    3.1      
u = mean =    4.3      
          
s = standard deviation =    1.2      
          
Thus,          
          
z = (x - u) / s =    -1      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1   ) =    0.158655254 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    1.9      
x2 = upper bound =    5      
u = mean =    4.3      
          
s = standard deviation =    1.2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    0.583333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.720165536      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.697415404      

Thus, those outside this interval is the complement =    0.302584596   [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.5      
          
Then, using table or technology,          
          
z =    0      
          
As x = u + z * s,          
          
where          
          
u = mean =    4.3      
z = the critical z score =    0      
s = standard deviation =    1.2      
          
Then          
          
x = critical value =    4.3   [ANSWER]