The lifetime of a light bulb in a certain application is nor

The lifetime of a light bulb in a certain application is normally distributed with a mean of 1,400 hours and standard deviation of 200 hours.

a ) What is the probability that a light bulb will last more than 1,800 hours?

b ) What lifetime is exceeded by 90% of light bulbs?

c ) What lifetime is exceeded by 20% of light bulbs?

d ) What is the probability that the lifetime of a light bulb is between 1,350 and 1,550 hours?

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1800      
u = mean =    1400      
          
s = standard deviation =    200      
          
Thus,          
          
z = (x - u) / s =    2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2   ) =    0.022750132 [ANSWER]

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B)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    1400      
z = the critical z score =    -1.281551566      
s = standard deviation =    200      
          
Then          
          
x = critical value =    1143.689687   [answer]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.8      
          
Then, using table or technology,          
          
z =    0.841621234      
          
As x = u + z * s,          
          
where          
          
u = mean =    1400      
z = the critical z score =    0.841621234      
s = standard deviation =    200      
          
Then          
          
x = critical value =    1568.324247   [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    1350      
x2 = upper bound =    1550      
u = mean =    1400      
          
s = standard deviation =    200      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.25      
z2 = upper z score = (x2 - u) / s =    0.75      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.401293674      
P(z < z2) =    0.773372648      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.372078973   [ANSWER]