Average talk time between charges of a cell phone is advertised as 5.2 hours. As

Question

Average talk time between charges of a cell phone is advertised as 5.2 hours. Assume that talk time is normally distributed with a standard deviation of 0.4 hour. Use Table 1.

a. Find the probability that talk time between charges for a randomly selected cell phone is below 4.9 hours. (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

                      Probability  _________

b. Find the probability that talk time between charges for a randomly selected cell phone is either more than 5.7 hours or below 3.7 hours. (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

                 Probability  ______________

c.Twenty six 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.)

                  Value  ____________

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    4.9      
u = mean =    5.2      
          
s = standard deviation =    0.4      
          
Thus,          
          
z = (x - u) / s =    -0.75      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.75   ) =    0.2266 [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 =    3.7      
x2 = upper bound =    5.7      
u = mean =    5.2      
          
s = standard deviation =    0.4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3.75      
z2 = upper z score = (x2 - u) / s =    1.25      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.0001      
P(z < z2) =    0.8944      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.8943      

Thus, those outside this interval is the complement =    0.1057   [ANSWER]  

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.26      
          
Then, using table or technology,          
          
z =    -0.64      
          
As x = u + z * s,          
          
where          
          
u = mean =    5.2      
z = the critical z score =    -0.64      
s = standard deviation =    0.4      
          
Then          
          
x = critical value =    4.944   [ANSWER]  

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