Suppose the life of a particular brand of calculator battery is approximately no

Question

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. (complete a-c)

A. What is the probability that a single battery randomly selected will have life between 70 and 80 hours? (please explain steps)

B. What is the probability that 9 randomly sampled batteries will have sample mean life between 70 and 80 hours ? (please explain steps)

B. If the manufacturer of the battery company is able to reduce the standard deviation of the battery from 10 to 8 hours, what would be the probability that 9 battery sampled will have a life between 70 and 80 hours?

Explanation / Answer

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    70      
x2 = upper bound =    80      
u = mean =    75      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.5      
z2 = upper z score = (x2 - u) / s =    0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.308537539      
P(z < z2) =    0.691462461      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.382924923   [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 =    70      
x2 = upper bound =    80      
u = mean =    75      
n = sample size =    9      
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.5      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.866385597   [answer]

*****************

c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    70      
x2 = upper bound =    80      
u = mean =    75      
n = sample size =    9      
s = standard deviation =    8      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.875      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.875      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.030396362      
P(z < z2) =    0.969603638      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.939207276   [answer]  
  
  

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