Let us assume that the battery life an iPod is nor mally dis

Let us assume that the battery life an iPod is nor- mally distributed with an average of 2.2 days and a standard deviation of 0.3 days. An iPod is selected at random and its battery life is tested.

(i). Calculate the probability that the battery life of the iPod is less than 1.8 days.

(ii). Calculate the probability that the battery life of the iPod is at least 2.7 days.

(iii). Calculate the probability that the battery life of the iPod is between 1.5 days and 2.4 days inclusively.

(iv).95.05% of the iPods have a battery life greater than how many days?

Solution

i.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1.8      
u = mean =    2.2      
          
s = standard deviation =    0.3      
          
Thus,          
          
z = (x - u) / s =    -1.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.333333333   ) =    0.09121122 [ANSWER]

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ii.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    2.7      
u = mean =    2.2      
          
s = standard deviation =    0.3      
          
Thus,          
          
z = (x - u) / s =    1.666666667      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.666666667   ) =    0.047790352 [ANSWER]

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iii.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    1.5      
x2 = upper bound =    2.4      
u = mean =    2.2      
          
s = standard deviation =    0.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.333333333      
z2 = upper z score = (x2 - u) / s =    0.666666667      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.009815329      
P(z < z2) =    0.747507462      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.737692134   [ANSWER]

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iv.

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.9505 =   0.0495      
          
Then, using table or technology,          
          
z =    -1.649721064      
          
As x = u + z * s,          
          
where          
          
u = mean =    2.2      
z = the critical z score =    -1.649721064      
s = standard deviation =    0.3      
          
Then          
          
x = critical value =    1.705083681   [ANSWER]