A light bulb manufacturer claims that, on average, their light bulbs burn for 30

Question

A light bulb manufacturer claims that, on average, their light bulbs burn for 3000 hours with a standard deviation of 1000 hours. You are the maintenance manager for Turner Field and you installed 400 of these light bulbs in one of the light standards at the ballpark. Assume the manufacturer's claim is correct. What is the probability that the average life time for these bulbs is greater than 3075 hours? What is the probability that the average life time for these bulbs is between 2960 hours and 3095 hours? You tracked the life times of all of the bulbs that you installed and determined that the mean life time for these bulbs was 2900 hours. What, if anything, does this tell you about the manufacturer's claim?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    3075      
u = mean =    3000      
n = sample size =    400      
s = standard deviation =    1000      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.5   ) =    0.066807201 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    2960      
x2 = upper bound =    3095      
u = mean =    3000      
n = sample size =    400      
s = standard deviation =    1000      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.8      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.9      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.211855399      
P(z < z2) =    0.97128344      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.759428042   [ANSWER]

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C)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2900      
u = mean =    3000      
n = sample size =    400      
s = standard deviation =    1000      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2   ) =    0.022750132

As this is a low probability (that the bulbs mean lifetime is 2900 hours or less), then there is reason to doubt the manufacturer's claima bout the mean lifetime of 3000 hours.
     

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